Question

How many grams of beryllium phosphate are produced when 38 grams of beryllium oxide reacts with iron (III) phosphate? Show your work.
3BeO + 2FePO₄ → Be₃(PO₄)₂ + Fe₂O₃

Answer 1

Answer: 305 g of $Be_3(PO_4)_2$ will be produced from 38 grams  of beryllium oxide

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$    

$\text{Moles of} BeO=\frac{38g}{25g/mol}=1.52moles$

$3BeO+2FePO_4\rightarrow Be_3(PO_4)_2+Fe_2O_3$

According to stoichiometry :

3 moles of $BeO$ produce = 1 mole of $Be_3(PO_4)_2$

Thus 1.52 moles of $BeO$ will produce =$\frac{1}{3}\times 1.52=0.507moles$  of $Be_3(PO_4)_2$

Mass of $Be_3(PO_4)_2=moles\times {\text {Molar mass}}=0.507moles\times 602g/mol=305g$

Thus 305 g of $Be_3(PO_4)_2$ will be produced from 38 grams  of beryllium oxide