Answer: 305 g of [tex]Be_3(PO_4)_2[/tex] will be produced from 38 grams of beryllium oxide
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} BeO=\frac{38g}{25g/mol}=1.52moles[/tex]
[tex]3BeO+2FePO_4\rightarrow Be_3(PO_4)_2+Fe_2O_3[/tex]
According to stoichiometry :
3 moles of [tex]BeO[/tex] produce = 1 mole of [tex]Be_3(PO_4)_2[/tex]
Thus 1.52 moles of [tex]BeO[/tex] will produce =[tex]\frac{1}{3}\times 1.52=0.507moles[/tex] of [tex]Be_3(PO_4)_2[/tex]
Mass of [tex]Be_3(PO_4)_2=moles\times {\text {Molar mass}}=0.507moles\times 602g/mol=305g[/tex]
Thus 305 g of [tex]Be_3(PO_4)_2[/tex] will be produced from 38 grams of beryllium oxide
