Answer:
1. The products of this reaction are ZnCl₂ and H₃PO₄.
2. 14.57 g.
Explanation:
<em>1. What would the products of this reaction be?</em>
- The balanced reaction between Zn₃(PO₄)₂ and HCl is represented as:
<em>Zn₃(PO₄)₂ + 6HCl → 3ZnCl₂ + 2H₃PO₄,</em>
It is clear that 1.0 mol of Zn₃(PO₄)₂ reacts with 6.0 mol of HCl to produce 3.0 mol of ZnCl₂ and 2.0 mol of H₃PO₄.
So, the products of this reaction are ZnCl₂ and H₃PO₄.
<em>2. If we produced 13.05 g of H₃PO₄, how many grams of hydrochloric acid would be need to start with?</em>
- Firstly, we should get the no. of moles (n) of 13.05 grams of H₃PO₄:
n = mass/molar mass = (13.05 g)/(97.994 g/mol) = 0.1332 mol.
<u><em>Using cross-multiplication:</em></u>
6.0 mol of HCl needed to produce → 2.0 mol of H₃PO₄, from stichiometry.
??? mol of HCl needed to produce → 0.1332 mol of H₃PO₄.
∴ The no. of moles of HCl needed = (6.0 mol)(0.1332 mol)/(2.0 mol) = 0.3995 mol.
∴ The mass of HCl needed = n*molar mass = (0.3995 mol)(36.46 g/mol) = 14.57 g.
<em>So, the grams of hydrochloric acid would be need to start with = 14.57 g.</em>
