To determine how many grams of [tex]\( \text{Zn}_{3}(\text{PO}_{4})_{2} \)[/tex] will be produced from 3.20 grams of [tex]\( (\text{NH}_{4})_{3}\text{PO}_{4} \)[/tex], we need to follow a series of steps involving stoichiometry and using the provided information.
1. Calculate the molar mass of [tex]\( (\text{NH}_{4})_{3}\text{PO}_{4} \)[/tex]:
- Molar mass of [tex]\( (\text{NH}_{4})_{3}\text{PO}_{4} \)[/tex] is [tex]\( 149.09 \)[/tex] g/mol.
2. Calculate the moles of [tex]\( (\text{NH}_{4})_{3}\text{PO}_{4} \)[/tex] given:
- Given mass of [tex]\( (\text{NH}_{4})_{3}\text{PO}_{4} \)[/tex] is [tex]\( 3.20 \)[/tex] grams.
- Moles of [tex]\( (\text{NH}_{4})_{3}\text{PO}_{4} \)[/tex] = [tex]\(\frac{3.20 \text{ g}}{149.09 \text{ g/mol}} \approx 0.0215 \)[/tex] moles.
3. Understand the stoichiometric ratio from the balanced equation:
- According to the equation: [tex]\( 2 (\text{NH}_{4})_{3}\text{PO}_{4} \rightarrow 1 \text{Zn}_{3}(\text{PO}_{4})_{2} \)[/tex],
- This means 2 moles of [tex]\( (\text{NH}_{4})_{3}\text{PO}_{4} \)[/tex] produces 1 mole of [tex]\( \text{Zn}_{3}(\text{PO}_{4})_{2} \)[/tex].
- The ratio is [tex]\( \frac{1}{2} \)[/tex].
4. Calculate the moles of [tex]\( \text{Zn}_{3}(\text{PO}_{4})_{2} \)[/tex] produced:
- Moles of [tex]\( \text{Zn}_{3}(\text{PO}_{4})_{2} \)[/tex] = [tex]\( 0.0215 \text{ moles of }(\text{NH}_{4})_{3}\text{PO}_{4} \times \frac{1}{2} \approx 0.0107 \)[/tex] moles.
5. Calculate the grams of [tex]\( \text{Zn}_{3}(\text{PO}_{4})_{2} \)[/tex]:
- Molar mass of [tex]\( \text{Zn}_{3}(\text{PO}_{4})_{2} \)[/tex] is [tex]\( 386.11 \)[/tex] g/mol.
- Grams of [tex]\( \text{Zn}_{3}(\text{PO}_{4})_{2} \)[/tex]:
[tex]\[ 0.0107 \text{ moles} \times 386.11 \text{ g/mol} \approx 4.14 \text{ grams}. \][/tex]
Therefore, the amount of [tex]\( \text{Zn}_{3}(\text{PO}_{4})_{2} \)[/tex] produced from 3.20 grams of [tex]\( (\text{NH}_{4})_{3}\text{PO}_{4} \)[/tex] is [tex]\( 4.14 \)[/tex] grams.
