Question

q.5. balance the following chemical equations.
a) alcl₃ + h₂o → al(oh)₃ + hcl
b) c₃h₈ + o₂ → co₂ + h₂o
c) al + feo → al₂o₃ + fe
d) s₈ + o₂ → __so₃
e) c₇h₆o₂ + o₂ → co₂ + h₂o
f) baf₂ + k₃po₄ → ba₃(po₄)₂ + kf
g) fecl₃ + naoh → fe(oh)₃ + nacl
h) k₃po₄ + hcl → kcl + h₃po₄
i) cabr₂ + na₃p → ca₃p₂ + nabr
j) b₂br₆ + hno₃ → b(no₃)₃ + hbr

Explanation:

Step1: Balance equation h

For the reaction $K_3PO_4 + HCl
ightarrow KCl + H_3PO_4$, balance potassium first. There are 3 potassium atoms in $K_3PO_4$, so we need 3 moles of $KCl$. Then, to balance chlorine, since we have 3 moles of $KCl$, we need 3 moles of $HCl$. The balanced equation is $1K_3PO_4+3HCl
ightarrow3KCl + 1H_3PO_4$.

Step2: Balance equation j

For the reaction $B_2Br_6+HNO_3
ightarrow B(NO_3)_3 + HBr$, balance boron first. There are 2 boron atoms in $B_2Br_6$, so we need 2 moles of $B(NO_3)_3$. Then, for the nitrate - ions, since we have 2 moles of $B(NO_3)_3$, we need 6 moles of $HNO_3$. Finally, for bromine, since we have 6 moles of bromine in $B_2Br_6$, we need 6 moles of $HBr$. The balanced equation is $1B_2Br_6 + 6HNO_3
ightarrow2B(NO_3)_3+6HBr$.

Answer:

a) $1AlCl_3 + 3H_2O
ightarrow1Al(OH)_3+3HCl$
b) $1C_3H_8 + 5O_2
ightarrow3CO_2 + 4H_2O$
c) $2Al+3FeO
ightarrow1Al_2O_3 + 3Fe$
d) $1S_8+12O_2
ightarrow8SO_3$
e) $2C_7H_6O_2+15O_2
ightarrow14CO_2+6H_2O$
f) $3BaF_2+2K_3PO_4
ightarrow1Ba_3(PO_4)_2 + 6KF$
g) $1FeCl_3+3NaOH
ightarrow1Fe(OH)_3+3NaCl$
h) $1K_3PO_4+3HCl
ightarrow3KCl + 1H_3PO_4$
i) $3CaBr_2+2Na_3P
ightarrow1Ca_3P_2+6NaBr$
j) $1B_2Br_6 + 6HNO_3
ightarrow2B(NO_3)_3+6HBr$