C5T Late submission until Today at 8 AM CST is allowed, with a one time 10% pena

Question

C5T Late submission until Today at 8 AM CST is allowed, with a one time 10% penalty to submitted score. Use the References to access important values if needed for this question. Manganese(III) oxide can be prepared by heating manganese(IV) oxide in vacuo at high temperature: The reaction of 471.8 g of MnO2 yields 351.3 g of Mn203. Calculate the theoretical yield of Mn 0, (assuming complete reaction) and its percentage yield Theoretical yield Percentage yield Submit Answer 5 question attempts remaining Back Autosaved at 5:27 AM

Explanation / Answer

moles of MnO2 = 471.8 /86.93 = 5.427 mol

so mole of Mn2O3 formed = 5.427/2 = 2.7135 mol

mass of Mn2O3 formed= mole x molar mass

= 2.7135 x 157.87 = 428.38 g

theoretical yield= 428.38 g

percentage yield = (351.3/428.38) x 100

= 80.007 %

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