To determine how many grams of Zn₃(PO₄)₂ will be produced by 0.50 grams of (NH₄)₃PO₄, follow these steps:
1. Calculate the molar mass of (NH₄)₃PO₄:
- Molar mass of NH₄ (Ammonium ion): [tex]\( 14.01 (N) + 4.04 (H) = 18.05 \)[/tex] g/mol
- Since there are 3 NH₄ groups: [tex]\( 3 \times 18.05 = 54.15 \)[/tex] g/mol
- Molar mass of P (Phosphorus): [tex]\( 30.97 \)[/tex] g/mol
- Molar mass of O (Oxygen): [tex]\( 16 \)[/tex] g/mol, and there are 4 oxygens: [tex]\( 4 \times 16 = 64 \)[/tex] g/mol
- Total molar mass of (NH₄)₃PO₄: [tex]\( 54.15 + 30.97 + 64 = 149.12 \)[/tex] g/mol
2. Calculate the moles of (NH₄)₃PO₄:
- Given mass of (NH₄)₃PO₄: [tex]\( 0.50 \)[/tex] g
- Moles of (NH₄)₃PO₄: [tex]\( \frac{0.50 \text{ g}}{149.12 \text{ g/mol}} = 0.003353 \)[/tex] mol
3. Determine the moles of Zn₃(PO₄)₂ produced:
- From the balanced chemical equation:
[tex]\( 3 \text{ ZnCl}_2 + 2 \text{ (NH}_4\text{)}_3 \text{PO}_4 \rightarrow \text{ Zn}_3(\text{PO}_4)_2 + 6 \text{ NH}_4 \text{Cl} \)[/tex]
- 2 moles of (NH₄)₃PO₄ produce 1 mole of Zn₃(PO₄)₂.
- Therefore, moles of Zn₃(PO₄)₂ produced: [tex]\( 0.003353 \times \frac{1}{2} = 0.001677 \)[/tex] mol
4. Calculate the molar mass of Zn₃(PO₄)₂:
- Molar mass of Zn (Zinc): [tex]\( 65.38 \)[/tex] g/mol, and there are 3 Zn atoms: [tex]\( 3 \times 65.38 = 196.14 \)[/tex] g/mol
- Molar mass of P (Phosphorus): [tex]\( 30.97 \)[/tex] g/mol, and there are 2 P atoms: [tex]\( 2 \times 30.97 = 61.94 \)[/tex] g/mol
- Molar mass of O (Oxygen): [tex]\( 16 \)[/tex] g/mol, and there are 8 O atoms: [tex]\( 8 \times 16 = 128 \)[/tex] g/mol
- Total molar mass of Zn₃(PO₄)₂: [tex]\( 196.14 + 61.94 + 128 = 386.08 \)[/tex] g/mol
5. Calculate the mass of Zn₃(PO₄)₂ produced:
- Moles of Zn₃(PO₄)₂: [tex]\( 0.001677 \)[/tex] mol
- Mass of Zn₃(PO₄)₂: [tex]\( 0.001677 \text{ mol} \times 386.08 \text{ g/mol} = 0.647 \)[/tex] g
Therefore, 0.50 grams of (NH₄)₃PO₄ will produce approximately 0.647 grams of Zn₃(PO₄)₂. The correct answer from the given options is:
[tex]\[ 0.65 \text{ g} \][/tex]
