Question

balancing chemical equations
balancing chemical equations drill
level 3 (advanced)
k₂o + h₂o → koh
cr + o₂ → cr₂o₃
cu(no₃)₂ + koh → cu(oh)₂ + kno₃
bacl₂ + na₂so₄ → nacl + baso₄
so₂ + o₂ → so₃
ni + s₈ → ni₂s₃
sc₂o₃ + h₂o → sc(oh)₃

Explanation:

Step1: Balance K, O, and H in $K_2O + H_2O

ightarrow KOH$
$K_2O + H_2O = 2KOH$

Step2: Balance Cr and O in $Cr+O_2

ightarrow Cr_2O_3$
$4Cr + 3O_2 = 2Cr_2O_3$

Step3: Balance Cu, N, O, K, and H in $Cu(NO_3)_2+KOH

ightarrow Cu(OH)_2 + KNO_3$
$Cu(NO_3)_2+2KOH = Cu(OH)_2+2KNO_3$

Step4: Balance Ba, Cl, Na, S, and O in $BaCl_2+Na_2SO_4

ightarrow NaCl + BaSO_4$
$BaCl_2+Na_2SO_4 = 2NaCl+BaSO_4$

Step5: Balance S and O in $SO_2+O_2

ightarrow SO_3$
$2SO_2+O_2 = 2SO_3$

Step6: Balance Ni and S in $Ni + S_8

ightarrow Ni_2S_3$
$16Ni + 3S_8 = 8Ni_2S_3$

Step7: Balance Sc, O, and H in $Sc_2O_3+H_2O

ightarrow Sc(OH)_3$
$Sc_2O_3+3H_2O = 2Sc(OH)_3$

Answer:

$K_2O + H_2O = 2KOH$; $4Cr + 3O_2 = 2Cr_2O_3$; $Cu(NO_3)_2+2KOH = Cu(OH)_2+2KNO_3$; $BaCl_2+Na_2SO_4 = 2NaCl+BaSO_4$; $2SO_2+O_2 = 2SO_3$; $16Ni + 3S_8 = 8Ni_2S_3$; $Sc_2O_3+3H_2O = 2Sc(OH)_3$