The mass of ferric nitrate, Fe(NO₃)₃ produced from the reaction given the data is 5792.44 g
Balanced equation
2Fe + 6HNO₃ –> 2Fe(NO₃)₃ + 3H₂
Molar mass of Fe = 56 g/mole
Mass of Fe from the balanced equation = 2 × 56 = 112 g
Molar mass of Fe(NO₃)₃ = 56 + 3[14 + (16×3)] = 242 g/mole
Mass of Fe(NO₃)₃ from the balanced equation = 2 × 242 = 484 g
SUMMARY
From the balanced equation above,
112 g of Fe reacted to produce 484 g of Fe(NO₃)₃
How to determine the mass of Fe(NO₃)₃ produced
From the balanced equation above,
112 g of Fe reacted to produce 484 g of Fe(NO₃)₃
Therefore,
1340.4 g of Fe will react to produce = (1340.4 × 484) / 112 = 5792.44 g of Fe(NO₃)₃
Thus, 5792.44 g of Fe(NO₃)₃ were obtained from the reaction
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