1) 1.56g of As2S3,0.140g of water,1.23g of nitric acid,and 3.50g of sodium nitrate are reacted to the following equation:
3As2S3+4H2O+10HNO3+18NaNO3--------------…
What mass of H3AsO4 is produced?
3As2S3+4H2O+10HNO3+18NaNO3--------------…
What mass of H3AsO4 is produced?
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3 As2S3 + 4 H2O + 10 HNO3 + 18 NaNO3 → 6 H3AsO4 + 28 NO + 9 Na2SO4
1.56g As2S3 / (246.0397 g/mol) = 0.00634044018 mol As2S3
0.140g H2O / (18.0153 g/mol) = 0.00777117228 mol H2O
1.23g HNO3 / (63.0130 g/mol) = 0.0195197816 mol HNO3
3.50g NaNO3 / (84.9948 g/mol) = 0.0411789898 mol NaNO3
0.00777117228 mol H2O would react completely with 0.00777117228 x (3/4) = 0.00582837921 mol As2S3, and 0.00777117228 x (10/4) = 0.0194279307 mol HNO3, and 0.00777117228 x (18/4) = 0.0349702753 mol NaNO3, but in all three cases there is more of the other reactant present, so H2O is the limiting reactant.
(0.00777117228 mol H2O) x (6/4) x (141.9432 g/mol) = 1.65 g H3AsO4
1.56g As2S3 / (246.0397 g/mol) = 0.00634044018 mol As2S3
0.140g H2O / (18.0153 g/mol) = 0.00777117228 mol H2O
1.23g HNO3 / (63.0130 g/mol) = 0.0195197816 mol HNO3
3.50g NaNO3 / (84.9948 g/mol) = 0.0411789898 mol NaNO3
0.00777117228 mol H2O would react completely with 0.00777117228 x (3/4) = 0.00582837921 mol As2S3, and 0.00777117228 x (10/4) = 0.0194279307 mol HNO3, and 0.00777117228 x (18/4) = 0.0349702753 mol NaNO3, but in all three cases there is more of the other reactant present, so H2O is the limiting reactant.
(0.00777117228 mol H2O) x (6/4) x (141.9432 g/mol) = 1.65 g H3AsO4