When 10.0g of NH3 reacts, the actual yield of N2 is 8.50 g. What is the percent yield? 4NH3(s)+6NOg)--5N28)+6H2...
1) Ammonia, NH3, reacts with molecular oxygen, O2, to form nitric oxide, NO, and water:4NH3(g) + 5O2(g) = 4NO (g) +6H2O(l)A. What is the limiting reactant and what is the theoretical yield of NO?B. What is the theoretical yield of H2O?C. How many grams of excess reagent will be left over?D. If the actual yield of NO had been 91 g, what would be the percent yield of the reaction
NH3(g) and CuO(s) react to form Cu(s), H2O(l) and N2(g). What is the theoretical yield of Cu(s) in moles if the percent yield of Cu(s) is 86.0% and 7.00 grams of Cu(s) forms?
11.26 Determine the theoretical yield of HF, in grams, when 49.8 g of NH3 reacts with 49.8 g of F2. What is the percent yield if 7.22 g of HF is actually formed during the reaction? 2NH3(g) + 5F2(8) N2F4(8) + 6HF(8)
What is the percent yield in a reaction between 8.73 g of N2 and 2.26 g of H2 in which 1.72 g of NH3 is formed? N2(g) + 3H2(g) → 2NH3(g)
How many grams of N2 are produced from 100.0 g of NH3 according to the following reaction? 4NH3 + 6NO ------ > 5N2 + 6H20 Calculate the mass (g) of H2O needed to produce 150 g of Mg(OH)2 according to the equation: Mg2N2 + 6H2O ----- > 3Mg(OH)2 + 2NH3
Nitrogen dioxide reacts with water to produce oxygen and ammonia: 4NO2(g)+6H2O(g)→7O2(g)+4NH3(g) How many grams of NH3 can be produced when 4.10 L of NO2 reacts at 385 ∘C and 735 mmHg ?
2)4NH3(9)302(g)6H20() +t 2 N2(g) TIS i blsiy loutoo srt tordt navi ( noitoon (a) 2 moles of NH3 and 3 moles of O2 will produce ? moles N2? Sdrs wo d bro asxs ni triszanq i sno ridw (b (b) 5.0 g of NH3 and 5.0 g of O2 will produce ? g of H20? How many H20 molecules?
6 1 point A sample of 7.16 g NH3 on oxidation produces 2.84 g of H20. Calculate the percent yield. Reaction: 4NH3 +502-4NO + 6H20 Use the following molar masses in your calculation: NH3 = 17.03 g/mol; H2O = 18.02 g/mol Do not type the percent symbol with your answer! Type your answer... Previous
9 1 point A sample of 8.18 g NH3 on oxidation produces 4.78 g of H20. Calculate the percent yield. Reaction: 4NH3 +502 4NO+6H20 Use the following molar masses in your calculation: NH3 = 17.03 g/mol; H20 = 18.02 g/mol Do not type the percent symbol with your answer! Type your answer... I
an experiment only formed 14 g of product (actual yield), thus the percent yield of product for that experiment is 1.4/2.0x100 - 70% % yield actual yield theoretical Wald X 100 You can rearrange this equation to calculate the actual yield if given the % and theoretical yields, or the theoretical yield if given the % and actual yields. Let's Practice: 1. Answer the following questions based on the reaction below (5pts) 2 NaCl + Mg0 Na2O + MgCl2 What...