2. How many grams of zinc will be formed if 32.0 g of copper reacts with zinc nitrate? Copper(I)nitrate is the other product.
1 answer:
Answer:
16.46 g.
Explanation:
- It is a stichiometry problem.
- We should write the balance equation of the mentioned chemical reaction:
<em>2Cu + Zn(NO₃)₂ → Zn + 2Cu(NO₃).</em>
- It is clear that 2.0 moles of Cu reacts with 1.0 mole of Zn(NO₃)₂ to produce 1.0 mole of Zn and 2.0 moles of Cu(NO₃).
- We need to calculate the number of moles of the reacted Cu (32.0 g) using the relation:
<em>n = mass / molar mass</em>
- The no. of moles of Cu = mass / atomic mass = (32.0 g) / (63.546 g/mol) = 0.503 mol.
<u><em>Using cross multiplication:</em></u>
2.0 moles of Cu produces → 1.0 mole of Zn, from the stichiometry.
0.503 mole of Cu produces → ??? mole of Zn.
- The no. of moles of Zn produced = (1.0 mol)(0.503 mol) / (2.0 mol) = 0.2517 mol.
∴ The grams of Zn produced = no. of moles x atomic mass of Zn = (0.2517 mol)(65.38 g/mol) = 16.46 g.
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