A 0.046 mol sample of water (molar mass: 18.01 g/mol) is reacted with excess of Mg3N2 in the following reaction: Mg3N2 + 3 H20 –→ 2 NH3 + 3 MgO In this experiment, 0.46 g of NH3 (molar mass: 17.1 g/mol) was obtained. What is the percent yield of the reaction? 81.3 % O A. 64.7 % O B. OC 71.5 96 77.5 % D. 87.7 % O E.

A 0.046 mol sample of water (molar mass: 18.01 g/mol) is reacted with excess of Mg3N2 in the following reaction: Mg3N2 + 3 H20 –→ 2 NH3 + 3 MgO In this experiment, 0.46 g of NH3 (molar mass: 17.1 g/mol) was obtained. What is the percent yield of the reaction? 81.3 % O A. 64.7 % O B. OC 71.5 96 77.5 % D. 87.7 % O E.

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

An excess of sodium carbonate, Na2CO3, in solution is added to a solution containing 17.87 g CaCl2. After performing the experiment, 12.51 g of calcium carbonate, CaCO3, is produced. Calculate the percent yield of this reaction.
Combining 0.254 mol Fe2O30.254 mol Fe2O3 with excess carbon produced 18.2 g Fe.18.2 g Fe. Fe2O3+3C⟶2Fe+3COFe2O3+3C⟶2Fe+3CO What is the actual yield of iron in moles? What is the theoretical yield of iron in moles? What is the percent yield?
Consider the following reaction: PCl3 + Cl2 -> PCl5. If the percent yield is 66.3%, what mass of PCl5 is actually formed when 61.3 g of Cl2 reacts with excess PCl3?
Each step in the following process has a yield of 60.0%. CH4 + 4Cl₂ CCl₂ + 4HCI CCl₂ + 2 HF CC1₂F₂ + 2HCl The CCI, formed in the first step is used as a reactant in the second step. If 5.00 mol CH4 reacts, what is the total amount of HCI produced? Assume that Cl₂ and HF are present in excess. moles HCl: mol
Consider the reaction between ammonium and oxygen: 4 NH3 ) + 5 02 () --> 4 NO ) + 6 H20() A chemist allows 10.8 g of NH3 and 5.30 g of O2 to react. When the reaction is finished, the chemist collects 3.00 g NO. Determine the percent yield for the reaction.
What is the theoretical yield of HCI if 30.0 g of BCI3 and excess H20 are reacted according to the following balanced reaction? BC13(g) +3 H20(1) + H3BO3(s) + 3HCI(g) The following molar masses are: HCI = 36.46 g/mol BCI3 = 117.17 g/mol H2O = 18.02 g/mol
Consider the following unbalanced reaction which has a percent yield of 33.70%. C6H12O6(s) + O2(g) →CO2(g) + H2O(g) Calculate the mass of glucose (180.156 g/mol) required to produce 34.38 grams of water (18.015 g/m Type your answ
Synthesis of disulfur diclhoride S8 + 4Cl2 —> 4 S2 Cl2 0.7797 mol S8 and 1.410 mol Cl2 are combined to make S2Cl2 1.What is the limiting reactant? 2.How much excess reactant will be left 3. How many moles of product can be made
Consider the following reaction: N2(g) + 3 H2(g) → 2 NH3(g)If 59.7 g of NH3 (MM = 17.031 g/mol) are collected when 57.83 g of N2 (MM = 28.0134 g/mol) react with 17.55 g H2 (MM = 2.1565 g/mole). The limiting reactant is? The theoretical yield is in grams NH3. The percent yield is
Aspirin (C,Hs04) may be synthesized by mixing salicylic acid (C,H,O;) with excess acetic anhydride (C4H6O3). When 2.0 g of salicylic acid is mixed with excess acetic anhydride, 1.7 g of aspirin is collected. What is the percent yield of this reaction? 42. The balanced equation is C;H,O3 + C,H,O3 → C,HgO4+ CH,O2 А. 98% В. 85% С. 77% D. 65 % Е. 55 % 110//1
Combining 0.349 mol Fe,O, with excess carbon produced 17.9 g Fe. Fe, 0, + 3C 2 Fe + 3 CO What is the actual yield of iron in moles? actual yield: mol What is the theoretical yield of iron in moles?
During a lab preparation of aspirin, a student used 3.10 g of salicylic acid and an excess amount of acetic anhydride for the reaction and obtained 2.95 g of product aspirin. What is the percentage yield of aspirin obtained by the student? (molar mass of salicylic acid= 138.12g/mol and molar mass of aspirin = 180.16g/mol) O 73.0 O 95.2 65.8 O68.4