To determine which of the given chemical equations would have a coefficient of 3 in front of any of the reactants when balanced, we need to analyze each equation one by one and balance them.
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Zn + HCl → ZnCl2 + H2:
- Balancing:
- We have 1 Zn on both sides.
- For Cl, we need 2 HCl to obtain 2 Cl for ZnCl2.
- Thus, we have: Zn + 2 HCl → ZnCl2 + H2
- Coefficient 2 appears in front of HCl, not 3.
- Balancing:
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H2SO4 + B(OH)3 → B2(SO4)3 + H2O:
- Balancing:
- We have 2 B atoms on the product side, so we need 2 B(OH)3 on the reactant side.
- We have 3 SO4, so we balance with 3 H2SO4.
- Balancing hydrogen gives: 3 H2SO4 + 2 B(OH)3 → B2(SO4)3 + 6 H2O
- The coefficient 3 appears in front of H2SO4.
- Balancing:
-
Fe + AgNO3 → Fe(NO3)2 + Ag:
- Balancing:
- Fe is 1 on both sides.
- We need 2 AgNO3 to obtain 2 Ag and get 2 NO3 for Fe(NO3)2.
- So we write: Fe + 2 AgNO3 → Fe(NO3)2 + 2 Ag
- Coefficient 2 appears in front of AgNO3, not 3.
- Balancing:
-
C2H4O2 + O2 → CO2 + H2O:
- Balancing:
- For C, 2 C from C2H4O2 means we need 2 CO2.
- For H, we have 4 H from C2H4O2 so we need 2 H2O.
- Balancing gives us: 1 C2H4O2 + 2 O2 → 2 CO2 + 2 H2O
- Coefficient 2 appears in front of O2, not 3.
- Balancing:
From our analysis, the only equation that results in the coefficient 3 in front of a reactant while balancing is:
H2SO4 + B(OH)3 → B2(SO4)3 + H2O