51.09g of NH3 are completely reacted with an excess of the other reactant. If the percent yield is found to be 90.68%, how many grams of H2O are actually produced?
2007-04-16 11:49:02 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
1. balance the equation (4 - 5 - 4 - 6)
2. work out moles of NH3 (51.09/17)
3. multiply by 1.5 and then by 0.9068
4. multiply by 18
2007-04-16 11:53:36 · answer #1 · answered by Gervald F 7 · 0⤊ 0⤋
You've got to have a balanced equation first. 4NH3 + 5O2 --> 4NO + 6H2O grams A --> moles A --> moles B --> grams B A) molNH3 = 85.15g/17.0g/mol ~ 5.0 mol NH3 mol O2 = 5.0 mol NH3 x 5 mol O2/4mol NH3 (from the coefficients in the balanced equation) = 6.25 mol O2 g O2 = 6.25mol x 32.0 g/mol = 200 g O2 B) follow the same procedure as in A.
2016-05-17 04:48:04 · answer #2 · answered by ? 3 · 0⤊ 0⤋