chemistry - stoichiomistry help please?

Question

a solution containing2-g of sodium sulfite reacts with 7 mL of phosphoric acid (H3PO4). the concentration of the acid solution is that there are 1.83g of H3PO4 per mL of solution.

a) Determine the mass of the excess reactant remaning at completion.
b) grams of water produced
c) Moles of sodium phosphate produced.
d) grams of sulfur dioxide produced.

Please explain how u get the answer.

Answer 1

The reaction is
3Na2SO3 + 2H3PO4 >> 2Na3PO4 + 3SO2 + 3H2O

MM(Na2SO3) = 126 g/mol
2g/ 126 g/mol= 0,0159 moles of Na2SO3

MM(H3PO4)=98 g/mol
1.83g /98 g/mol= 0.0187 moles in 1 mL>> in 7 mL there are 0.131 moles

The ratio between Na2SO3 and H3PO4 is 3 : 2
so for 0.0159 moles of Na2SO3 we need 0.0106 moles of H3PO4 leaving 0.131-0.0106=0.120 moles H3PO4 in excess or 11.76 g

moles of water produced = 0.0159 or 0.286 g

moles of Na3PO4 =0.0159 (2/3)=0.0106

moles of SO2 produced = 0.0159 or 1.02g