1. If 6.91g are obtained froma reaction of 6.54 g Zn with excess of oxygen. What is the percent yield of the reaction? Molar mass: Zn =65.38 g/mol ,0=16.00 g/mol

Md Ariful Islam

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2Zn+O2--------2ZnO

To calculate the percent yield we must calculate the theoretical yield of ZnO first

Using a mass-mass relationship,

Molar mass of zinc=65.38, but since we have 2 moles it will be 2×65.38=130.76g

Molar mass of ZnO =81.38, but since there are aldo 2 moles ,we have 2×81.38=162.76

So if 130.76g of Zn yields 162.76g of ZnO

Then 6.54g of Zn will yield;

6.54×162.76/130.76=8.14g of ZnO

8.14g is the theoretical yield of ZnO

So to calculate percent yield we have:

(Actual yield/theoretical yield)×100%

Percent yield =(6.91/8.14)×100%

=84.9% which is approximately 85%

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