balancing equations worksheet\n1) h₃po₄ + koh → k₃po₄ + h₂o\n2) k + b₂o₃ → k₂o + …

balancing equations worksheet\n1) __ h₃po₄ + __ koh → __ k₃po₄ + __ h₂o\n2) __ k + __ b₂o₃ → __ k₂o + __ b\n3) __ hcl + __ naoh → __ nacl + __ h₂o\n4) __ na + __ nano₃ → __ na₂o + __ n₂\n5) __ c + __ s₈ → __ cs₂\n6) __ na + __ o₂ → __ na₂o\n7) __ n₂ + __ o₂ → __ n₂o₅\n8) __ h₃po₄ + __ mg(oh)₂ → __ mg₃(po₄)₂ + __ h₂o\n9) __ naoh + __ h₂co₃ → __ na₂co₃ + __ h₂o\n10) __ koh + __ hbr → __ kbr + __ h₂o\n11) __ na + __ o₂ → __ na₂o\n12) __ al(oh)₃ + __ h₂co₃ → __ al₂(co₃)₃ + __ h₂o\n13) __ al + __ s₈ → __ al₂s₃\n14) __ cs + __ n₂ → __ cs₃n\n15) __ mg + __ cl₂ → __ mgcl₂\n16) __ rb + __ rbno₃ → __ rb₂o + __ n₂\n17) __ c₆h₆ + __ o₂ → __ co₂ + __ h₂o\n18) __ n₂ + __ h₂ → __ nh₃\n19) __ c₁₀h₂₂ + __ o₂ → __ co₂ + __ h₂o\n20) __ al(oh)₃ + __ hbr → __ albr₃ + __ h₂o\n21) __ ch₃ch₂ch₂ch₃ + __ o₂ → __ co₂ + __ h₂o\n22) __ c₃h₈ + __ o₂ → __ co₂ + __ h₂o\n23) __ li + __ alcl₃ → __ licl + __ al\n24) __ c₂h₆ + __ o₂ → __ co₂ + __ h₂o\n25) __ nh₄oh + __ h₃po₄ → __ (nh₄)₃po₄ + __ h₂o\n26) __ rb + __ p → __ rb₃p\n27) __ ch₄ + __ o₂ → __ co₂ + __ h₂o\n28) __ al(oh)₃ + __ h₂so₄ → __ al₂(so₄)₃ + __ h₂o\n29) __ na + __ cl₂ → __ nacl\n30) __ rb + __ s₈ → __ rb₂s\n31) __ h₃po₄ + __ ca(oh)₂ → __ ca₃(po₄)₂ + __ h₂o\n32) __ nh₃ + __ hcl → __ nh₄cl\n33) __ li + __ h₂o → __ lioh + __ h₂\n34) __ ca₃(po₄)₂ + __ sio₂ + __ c → __ casio₃ + __ co + __ p\n35) __ nh₃ + __ o₂ → __ n₂ + __ h₂o\n36) __ fes₂ + __ o₂ → __ fe₂o₃ + __ so₂\n37) __ c + __ so₂ → __ cs₂ + __ co\neverett community college tutoring center

Answer

Explanation:

Step1: Balance equation 1

For the reaction $\mathrm{H_3PO_4}+\mathrm{KOH}\to\mathrm{K_3PO_4}+\mathrm{H_2O}$, balance potassium first. Since there are 3 potassium atoms in $\mathrm{K_3PO_4}$, we need 3 moles of $\mathrm{KOH}$. Then balance hydrogen and oxygen. The balanced equation is $\mathrm{H_3PO_4}+3\mathrm{KOH}\to\mathrm{K_3PO_4}+3\mathrm{H_2O}$.

Step2: Balance equation 2

For $\mathrm{K}+\mathrm{B_2O_3}\to\mathrm{K_2O}+\mathrm{B}$, balance potassium. We need 6 moles of $\mathrm{K}$ to get 3 moles of $\mathrm{K_2O}$. Then balance boron. The balanced equation is $6\mathrm{K}+\mathrm{B_2O_3}\to3\mathrm{K_2O}+2\mathrm{B}$.

Step3: Balance equation 3

For $\mathrm{HCl}+\mathrm{NaOH}\to\mathrm{NaCl}+\mathrm{H_2O}$, it is a simple acid - base neutralization. The balanced equation is $\mathrm{HCl}+\mathrm{NaOH}\to\mathrm{NaCl}+\mathrm{H_2O}$ (1:1:1:1 ratio).

Step4: Balance equation 4

For $\mathrm{Na}+\mathrm{NaNO_3}\to\mathrm{Na_2O}+\mathrm{N_2}$, balance nitrogen first. We need 10 moles of $\mathrm{Na}$ and 2 moles of $\mathrm{NaNO_3}$ to balance all elements. The balanced equation is $10\mathrm{Na}+2\mathrm{NaNO_3}\to6\mathrm{Na_2O}+\mathrm{N_2}$.

Step5: Balance equation 5

For $\mathrm{C}+\mathrm{S_8}\to\mathrm{CS_2}$, balance sulfur. We need 4 moles of $\mathrm{C}$ to react with 1 mole of $\mathrm{S_8}$ to form 4 moles of $\mathrm{CS_2}$. The balanced equation is $4\mathrm{C}+\mathrm{S_8}\to4\mathrm{CS_2}$.

Step6: Balance equation 6

For $\mathrm{Na}+\mathrm{O_2}\to\mathrm{Na_2O}$, balance oxygen. We need 4 moles of $\mathrm{Na}$ to react with 1 mole of $\mathrm{O_2}$ to form 2 moles of $\mathrm{Na_2O}$. The balanced equation is $4\mathrm{Na}+\mathrm{O_2}\to2\mathrm{Na_2O}$.

Step7: Balance equation 7

For $\mathrm{N_2}+\mathrm{O_2}\to\mathrm{N_2O_5}$, balance oxygen. We need 2 moles of $\mathrm{N_2}$ and 5 moles of $\mathrm{O_2}$ to form 2 moles of $\mathrm{N_2O_5}$. The balanced equation is $2\mathrm{N_2}+5\mathrm{O_2}\to2\mathrm{N_2O_5}$.

Step8: Balance equation 8

For $\mathrm{H_3PO_4}+\mathrm{Mg(OH)_2}\to\mathrm{Mg_3(PO_4)_2}+\mathrm{H_2O}$, balance magnesium first. We need 3 moles of $\mathrm{Mg(OH)_2}$ and 2 moles of $\mathrm{H_3PO_4}$. Then balance hydrogen and oxygen. The balanced equation is $2\mathrm{H_3PO_4}+3\mathrm{Mg(OH)_2}\to\mathrm{Mg_3(PO_4)_2}+6\mathrm{H_2O}$.

Step9: Balance equation 9

For $\mathrm{NaOH}+\mathrm{H_2CO_3}\to\mathrm{Na_2CO_3}+\mathrm{H_2O}$, balance sodium. We need 2 moles of $\mathrm{NaOH}$. The balanced equation is $2\mathrm{NaOH}+\mathrm{H_2CO_3}\to\mathrm{Na_2CO_3}+2\mathrm{H_2O}$.

Step10: Balance equation 10

For $\mathrm{KOH}+\mathrm{HBr}\to\mathrm{KBr}+\mathrm{H_2O}$, it is a simple acid - base reaction. The balanced equation is $\mathrm{KOH}+\mathrm{HBr}\to\mathrm{KBr}+\mathrm{H_2O}$.

Step11: Same as equation 6, $4\mathrm{Na}+\mathrm{O_2}\to2\mathrm{Na_2O}$.

Step12: For $\mathrm{Al(OH)_3}+\mathrm{H_2CO_3}\to\mathrm{Al_2(CO_3)_3}+\mathrm{H_2O}$, balance aluminum first. We need 2 moles of $\mathrm{Al(OH)_3}$ and 3 moles of $\mathrm{H_2CO_3}$. Then balance hydrogen and oxygen. The balanced equation is $2\mathrm{Al(OH)_3}+3\mathrm{H_2CO_3}\to\mathrm{Al_2(CO_3)_3}+6\mathrm{H_2O}$.

Step13: Given as balanced: $16\mathrm{Al}+3\mathrm{S_8}\to8\mathrm{Al_2S_3}$.

Step14: For $\mathrm{Cs}+\mathrm{N_2}\to\mathrm{Cs_3N}$, balance nitrogen first. We need 6 moles of $\mathrm{Cs}$ to react with 1 mole of $\mathrm{N_2}$ to form 2 moles of $\mathrm{Cs_3N}$. The balanced equation is $6\mathrm{Cs}+\mathrm{N_2}\to2\mathrm{Cs_3N}$.

Step15: For $\mathrm{Mg}+\mathrm{Cl_2}\to\mathrm{MgCl_2}$, it is already balanced ($1:1:1$ ratio).

Step16: For $\mathrm{Rb}+\mathrm{RbNO_3}\to\mathrm{Rb_2O}+\mathrm{N_2}$, balance nitrogen first. We need 10 moles of $\mathrm{Rb}$ and 2 moles of $\mathrm{RbNO_3}$ to balance all elements. The balanced equation is $10\mathrm{Rb}+2\mathrm{RbNO_3}\to6\mathrm{Rb_2O}+\mathrm{N_2}$.

Step17: For $\mathrm{C_6H_6}+\mathrm{O_2}\to\mathrm{CO_2}+\mathrm{H_2O}$, balance carbon first. We need 2 moles of $\mathrm{C_6H_6}$ and 15 moles of $\mathrm{O_2}$ to form 12 moles of $\mathrm{CO_2}$ and 6 moles of $\mathrm{H_2O}$. The balanced equation is $2\mathrm{C_6H_6}+15\mathrm{O_2}\to12\mathrm{CO_2}+6\mathrm{H_2O}$.

Step18: For $\mathrm{N_2}+\mathrm{H_2}\to\mathrm{NH_3}$, balance nitrogen first. We need 1 mole of $\mathrm{N_2}$ and 3 moles of $\mathrm{H_2}$ to form 2 moles of $\mathrm{NH_3}$. The balanced equation is $\mathrm{N_2}+3\mathrm{H_2}\to2\mathrm{NH_3}$.

Step19: For $\mathrm{C_{10}H_{22}}+\mathrm{O_2}\to\mathrm{CO_2}+\mathrm{H_2O}$, balance carbon first. We need 2 moles of $\mathrm{C_{10}H_{22}}$ and 31 moles of $\mathrm{O_2}$ to form 20 moles of $\mathrm{CO_2}$ and 22 moles of $\mathrm{H_2O}$. The balanced equation is $2\mathrm{C_{10}H_{22}}+31\mathrm{O_2}\to20\mathrm{CO_2}+22\mathrm{H_2O}$.

Step20: For $\mathrm{Al(OH)_3}+\mathrm{HBr}\to\mathrm{AlBr_3}+\mathrm{H_2O}$, balance bromine first. We need 3 moles of $\mathrm{HBr}$ for 1 mole of $\mathrm{Al(OH)_3}$. The balanced equation is $\mathrm{Al(OH)_3}+3\mathrm{HBr}\to\mathrm{AlBr_3}+3\mathrm{H_2O}$.

Step21: Given as balanced: $2\mathrm{CH_3CH_2CH_2CH_3}+13\mathrm{O_2}\to8\mathrm{CO_2}+10\mathrm{H_2O}$.

Step22: For $\mathrm{C_3H_8}+\mathrm{O_2}\to\mathrm{CO_2}+\mathrm{H_2O}$, balance carbon first. We need 1 mole of $\mathrm{C_3H_8}$ and 5 moles of $\mathrm{O_2}$ to form 3 moles of $\mathrm{CO_2}$ and 4 moles of $\mathrm{H_2O}$. The balanced equation is $\mathrm{C_3H_8}+5\mathrm{O_2}\to3\mathrm{CO_2}+4\mathrm{H_2O}$.

Step23: For $\mathrm{Li}+\mathrm{AlCl_3}\to\mathrm{LiCl}+\mathrm{Al}$, balance chlorine first. We need 3 moles of $\mathrm{Li}$ to react with 1 mole of $\mathrm{AlCl_3}$ to form 3 moles of $\mathrm{LiCl}$ and 1 mole of $\mathrm{Al}$. The balanced equation is $3\mathrm{Li}+\mathrm{AlCl_3}\to3\mathrm{LiCl}+\mathrm{Al}$.

Step24: For $\mathrm{C_2H_6}+\mathrm{O_2}\to\mathrm{CO_2}+\mathrm{H_2O}$, balance carbon first. We need 2 moles of $\mathrm{C_2H_6}$ and 7 moles of $\mathrm{O_2}$ to form 4 moles of $\mathrm{CO_2}$ and 6 moles of $\mathrm{H_2O}$. The balanced equation is $2\mathrm{C_2H_6}+7\mathrm{O_2}\to4\mathrm{CO_2}+6\mathrm{H_2O}$.

Step25: For $\mathrm{NH_4OH}+\mathrm{H_3PO_4}\to(\mathrm{NH_4})_3\mathrm{PO_4}+\mathrm{H_2O}$, balance ammonium first. We need 3 moles of $\mathrm{NH_4OH}$. The balanced equation is $3\mathrm{NH_4OH}+\mathrm{H_3PO_4}\to(\mathrm{NH_4})_3\mathrm{PO_4}+3\mathrm{H_2O}$.

Step26: For $\mathrm{Rb}+\mathrm{P}\to\mathrm{Rb_3P}$, balance rubidium first. We need 3 moles of $\mathrm{Rb}$. The balanced equation is $3\mathrm{Rb}+\mathrm{P}\to\mathrm{Rb_3P}$.

Step27: For $\mathrm{CH_4}+\mathrm{O_2}\to\mathrm{CO_2}+\mathrm{H_2O}$, balance carbon first. We need 1 mole of $\mathrm{CH_4}$ and 2 moles of $\mathrm{O_2}$ to form 1 mole of $\mathrm{CO_2}$ and 2 moles of $\mathrm{H_2O}$. The balanced equation is $\mathrm{CH_4}+2\mathrm{O_2}\to\mathrm{CO_2}+2\mathrm{H_2O}$.

Step28: For $\mathrm{Al(OH)_3}+\mathrm{H_2SO_4}\to\mathrm{Al_2(SO_4)_3}+\mathrm{H_2O}$, balance aluminum first. We need 2 moles of $\mathrm{Al(OH)_3}$ and 3 moles of $\mathrm{H_2SO_4}$. Then balance hydrogen and oxygen. The balanced equation is $2\mathrm{Al(OH)_3}+3\mathrm{H_2SO_4}\to\mathrm{Al_2(SO_4)_3}+6\mathrm{H_2O}$.

Step29: For $\mathrm{Na}+\mathrm{Cl_2}\to\mathrm{NaCl}$, balance chlorine first. We need 2 moles of $\mathrm{Na}$ to react with 1 mole of $\mathrm{Cl_2}$ to form 2 moles of $\mathrm{NaCl}$. The balanced equation is $2\mathrm{Na}+\mathrm{Cl_2}\to2\mathrm{NaCl}$.

Step30: Given as balanced: $16\mathrm{Rb}+1\mathrm{S_8}\to8\mathrm{Rb_2S}$.

Step31: For $\mathrm{H_3PO_4}+\mathrm{Ca(OH)_2}\to\mathrm{Ca_3(PO_4)_2}+\mathrm{H_2O}$, balance calcium first. We need 3 moles of $\mathrm{Ca(OH)_2}$ and 2 moles of $\mathrm{H_3PO_4}$. Then balance hydrogen and oxygen. The balanced equation is $2\mathrm{H_3PO_4}+3\mathrm{Ca(OH)_2}\to\mathrm{Ca_3(PO_4)_2}+6\mathrm{H_2O}$.

Step32: For $\mathrm{NH_3}+\mathrm{HCl}\to\mathrm{NH_4Cl}$, it is already balanced ($1:1:1$ ratio).

Step33: For $\mathrm{Li}+\mathrm{H_2O}\to\mathrm{LiOH}+\mathrm{H_2}$, balance hydrogen first. We need 2 moles of $\mathrm{Li}$ and 2 moles of $\mathrm{H_2O}$ to form 2 moles of $\mathrm{LiOH}$ and 1 mole of $\mathrm{H_2}$. The balanced equation is $2\mathrm{Li}+2\mathrm{H_2O}\to2\mathrm{LiOH}+\mathrm{H_2}$.

Step34: For $\mathrm{Ca_3(PO_4)_2}+\mathrm{SiO_2}+\mathrm{C}\to\mathrm{CaSiO_3}+\mathrm{CO}+\mathrm{P}$, balance calcium first. The balanced equation is $\mathrm{Ca_3(PO_4)_2}+3\mathrm{SiO_2}+5\mathrm{C}\to3\mathrm{CaSiO_3}+5\mathrm{CO}+2\mathrm{P}$.

Step35: For $\mathrm{NH_3}+\mathrm{O_2}\to\mathrm{N_2}+\mathrm{H_2O}$, balance nitrogen first. We need 4 moles of $\mathrm{NH_3}$ and 3 moles of $\mathrm{O_2}$ to form 2 moles of $\mathrm{N_2}$ and 6 moles of $\mathrm{H_2O}$. The balanced equation is $4\mathrm{NH_3}+3\mathrm{O_2}\to2\mathrm{N_2}+6\mathrm{H_2O}$.

Step36: For $\mathrm{FeS_2}+\mathrm{O_2}\to\mathrm{Fe_2O_3}+\mathrm{SO_2}$, balance iron first. We need 4 moles of $\mathrm{FeS_2}$ and 11 moles of $\mathrm{O_2}$ to form 2 moles of $\mathrm{Fe_2O_3}$ and 8 moles of $\mathrm{SO_2}$. The balanced equation is $4\mathrm{FeS_2}+11\mathrm{O_2}\to2\mathrm{Fe_2O_3}+8\mathrm{SO_2}$.

Step37: For $\mathrm{C}+\mathrm{SO_2}\to\mathrm{CS_2}+\mathrm{CO}$, balance sulfur first. We need 5 moles of $\mathrm{C}$ and 2 moles of $\mathrm{SO_2}$ to form 1 mole of $\mathrm{CS_2}$ and 4 moles of $\mathrm{CO}$. The balanced equation is $5\mathrm{C}+2\mathrm{SO_2}\to\mathrm{CS_2}+4\mathrm{CO}$.

Answer:

$1\mathrm{H_3PO_4}+3\mathrm{KOH}\to1\mathrm{K_3PO_4}+3\mathrm{H_2O}$
$6\mathrm{K}+1\mathrm{B_2O_3}\to3\mathrm{K_2O}+2\mathrm{B}$
$1\mathrm{HCl}+1\mathrm{NaOH}\to1\mathrm{NaCl}+1\mathrm{H_2O}$
$10\mathrm{Na}+2\mathrm{NaNO_3}\to6\mathrm{Na_2O}+1\mathrm{N_2}$
$4\mathrm{C}+1\mathrm{S_8}\to4\mathrm{CS_2}$
$4\mathrm{Na}+1\mathrm{O_2}\to2\mathrm{Na_2O}$
$2\mathrm{N_2}+5\mathrm{O_2}\to2\mathrm{N_2O_5}$
$2\mathrm{H_3PO_4}+3\mathrm{Mg(OH)_2}\to1\mathrm{Mg_3(PO_4)_2}+6\mathrm{H_2O}$
$2\mathrm{NaOH}+1\mathrm{H_2CO_3}\to1\mathrm{Na_2CO_3}+2\mathrm{H_2O}$
$1\mathrm{KOH}+1\mathrm{HBr}\to1\mathrm{KBr}+1\mathrm{H_2O}$
$4\mathrm{Na}+1\mathrm{O_2}\

balancing equations worksheet\n1) __ h₃po₄ + __ koh → __ k₃po₄ + __ h₂o\n2) __ k + __ b₂o₃ → __ k₂o + __…