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

Step-by-step explanation:

To calculate the moles :

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

`\text{Moles of} BeO=(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 =
`(1)/(3)* 1.52=0.507moles` of
`Be_3(PO_4)_2`

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

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