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For the reaction shown, compute the theoretical yield of product (in moles) for each of the following initial amounts of reactants.
For the reaction shown, compute the theoretical yield of product (in moles) for each of the following initial amounts of reactants.Mn(s)+O2(g)→MnO2(s)
a) 5 mol Mn
, 5 mol O2
b) 5 mol Mn
, 8 mol O2
c) 28.5 mol Mn
, 44.6 mol O2
(Express your answers using one significant figure)
Thetheoretical yieldof product (in moles) for each of the following initial amounts of reactants.a) 5 mol Mn, 5 mol O₂ : 5 mol MnO₂b) 5 mol Mn, 8 mol O₂ : 5 mol MnO₂c) 28.5 mol Mn, 44.6 mol O₂ : 28.5 mol MnO₂For the balanced reaction: 2 Mn(s) + O₂(g) → 2 MnO₂(s)a) 5 mol Mn, 5 mol O₂:Using thestoichiometryof the balanced reaction, 2 moles of Mn react with 1 mole of O₂. To determine thelimiting reactant, we calculate the ratio for each reactant:Mn: 5 mol / 2 = 2.5O₂: 5 mol / 1 = 5Mn is the limiting reactant. Thus, the theoretical yield of MnO₂ is:2.5 * 2 = 5 mol MnO₂mol MnO₂ = mol Mn = 5 molb) 5 mol Mn, 8 mol O₂:Mn: 5 mol / 2 = 2.5O₂: 8 mol / 1 = 8Mn is the limiting reactant. Thus, the theoretical yield of MnO₂ is:2.5 * 2 = 5 mol MnO₂mol MnO₂ = mol Mn = 5 molc) 28.5 mol Mn, 44.6 mol O₂:Mn: 28.5 mol / 2 = 14.25O₂: 44.6 mol / 1 = 44.6Mn is the limiting reactant. Thus, the theoretical yield of MnO₂ is:14.25 * 2 = 28.5 mol MnO₂mol MnO₂ = mol Mn = 28.5 molLearn more abouttheoretical yieldhere:brainly.com/question/25996347#SPJ11...