Answer:
[tex]Mole of Na_3 PO_4 = 0.092[/tex]
Explanation:
Firstly balance the chemical reaction,
[tex]3NaBr+H_3 PO_4[/tex] → [tex]Na_3 PO_4+3HBr[/tex]
Molecular weight of [tex]NaBr[/tex]=103g/mole;
Molecular weight of [tex]Na_3 PO_4[/tex] =164g/mole;
Moles are:
Number of mole of [tex]NaBr[/tex]=0.275
From the balanced equation,
3 mole of [tex]NaBr[/tex] gives 1 mole of [tex]Na_3 PO_4;[/tex]
1 mole of [tex]NaBr[/tex] gives 1/3 mole of [tex]Na_3 PO_4;[/tex]
Hence;
0.275 mole of [tex]NaBr[/tex] will produce [tex]\frac{0.275}{3}[/tex] mole of [tex]Na_3 PO_4;[/tex]
Hence,
[tex]Mole of Na_3 PO_4 = 0.092[/tex]
