balancing chemical equations\nbalancing chemical equations drill\nlevel 3 (advanced)\nk₂o + h₂o → koh\ncr +…
balancing chemical equations\nbalancing chemical equations drill\nlevel 3 (advanced)\nk₂o + h₂o → koh\ncr + o₂ → cr₂o₃\ncu(no₃)₂ + koh → cu(oh)₂ + kno₃\nbacl₂ + na₂so₄ → nacl + baso₄\nso₂ + o₂ → so₃\nni + s₈ → ni₂s₃\nsc₂o₃ + h₂o → sc(oh)₃
Answer
Explanation:
Step1: Balance K, O, and H in $K_2O + H_2O\rightarrow KOH$
$K_2O + H_2O = 2KOH$
Step2: Balance Cr and O in $Cr+O_2\rightarrow Cr_2O_3$
$4Cr + 3O_2 = 2Cr_2O_3$
Step3: Balance Cu, N, O, K, and H in $Cu(NO_3)_2+KOH\rightarrow Cu(OH)_2 + KNO_3$
$Cu(NO_3)_2+2KOH = Cu(OH)_2+2KNO_3$
Step4: Balance Ba, Cl, Na, S, and O in $BaCl_2+Na_2SO_4\rightarrow NaCl + BaSO_4$
$BaCl_2+Na_2SO_4 = 2NaCl+BaSO_4$
Step5: Balance S and O in $SO_2+O_2\rightarrow SO_3$
$2SO_2+O_2 = 2SO_3$
Step6: Balance Ni and S in $Ni + S_8\rightarrow Ni_2S_3$
$16Ni + 3S_8 = 8Ni_2S_3$
Step7: Balance Sc, O, and H in $Sc_2O_3+H_2O\rightarrow Sc(OH)_3$
$Sc_2O_3+3H_2O = 2Sc(OH)_3$
Answer:
$K_2O + H_2O = 2KOH$; $4Cr + 3O_2 = 2Cr_2O_3$; $Cu(NO_3)_2+2KOH = Cu(OH)_2+2KNO_3$; $BaCl_2+Na_2SO_4 = 2NaCl+BaSO_4$; $2SO_2+O_2 = 2SO_3$; $16Ni + 3S_8 = 8Ni_2S_3$; $Sc_2O_3+3H_2O = 2Sc(OH)_3$