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For each of the following reactions, 20.0 g of each
Chapter 6, Problem 71QP(choose chapter or problem)
For each of the following reactions, 20.0 g of each reactant is present initially. Determine the limiting reactant, and calculate the grams of product in parentheses that would be produced.a. 2Al(s) + 3Cl2(g) ?2AlCl3(s) (AlCl3)b. 4NH3(g) + 5O2(g) ?4NO(g) + 6H2O(g) (H2O)c. CS2(g) + 3O2(g) ?CO2(g) + 2SO2(g) (SO2)
Questions & Answers
QUESTION:
For each of the following reactions, 20.0 g of each reactant is present initially. Determine the limiting reactant, and calculate the grams of product in parentheses that would be produced.a. 2Al(s) + 3Cl2(g) ?2AlCl3(s) (AlCl3)b. 4NH3(g) + 5O2(g) ?4NO(g) + 6H2O(g) (H2O)c. CS2(g) + 3O2(g) ?CO2(g) + 2SO2(g) (SO2)
ANSWER:Solution 71 QPHere we have to determine the limiting reactant, and calculate the grams of product in parentheses that would be produced. For each of the following reactions, 20.0 g of each reactant is present initially. In this question as we have to calculate the mass of product in gram and mass of reactant is also given in gram, 1st we have to convert gram of each reactant into mole, then have to calculate moles of product and then finally have to express the amount of product from in gram. Step-1a. 2Al(s) + 3Cl2(g) 2AlCl3(s) (AlCl3)In this reaction 2 mole of AlCl3 has formed from 2 mole of Al and 2 moles of AlCl3 has formed from 3 moles of Cl2.1 mole of Al = 26.99 g/mole, than 2 mole of Al = 2 mole26.98 g/mole = 53.96 g 1 mole of Cl2 = 70.90 g/mol, then 3 moles of Cl = 3 mole70.90g/mol = 212.7 gMolecular mass of AlCl3 = 133.34 g/mol, 2 moles of AlCl3 = 2 moles133.34 g/mol = 266.68 gMass of AlCl3 produc