chemistry - stoichiomistry help please?

Question

a solution containing 2g 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 balanced reaction is:

3Na2SO3 + 2H3PO4 ----> 2Na3PO4 + 3SO2 + 3H2O

2 grams Na2SO3/126 g/mol = 0.0159 moles
7 ml H3PO4 * 1.83 g/ml * 1mole/98 g = 0.131 moles

a) Given the stoichiometry, H3PO4 is the excess reagent. With 0.0159 moles of sodium sulfate, you'll need 0.0106 moles of H3PO4, leaving 0.120 moles in excess, or 11.76 grams in excess.
b) You'll produce 0.0159 moles of water, or 0.286 grams.
c) You will produce 0.0159 * 2/3 or 0.0106 moles of Na3PO4
d) 0.0159 moles of SO2, or 1.02 grams.