A 115.0-g sample of oxygen was produced by heating 400.0 g of potassium chlorate.
2KClO3 Right arrow. 2KCI + 3O2
What is the percent yield of oxygen in this chemical reaction?
Use Percent yield equals StartFraction actual yield over theoretical yield EndFraction times 100..
69.63%
73.40%
90.82%
136.2%
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Molar mass of KClO3 is 122.5
2 moles of KClO3 has mass 122.5 * 2 = 245g
3 moles of O2 has mass = 3*32 = 96g
So, 245g of KCLO3 produce = 96g of O2
400g will produce = (96* 400)/245 = 156.7g of O2
Theoretical yield = 115 g (given)
%yield = (115/156.7)* 100 = 73.39% = 73.40%
2 moles of KClO3 has mass 122.5 * 2 = 245g
3 moles of O2 has mass = 3*32 = 96g
So, 245g of KCLO3 produce = 96g of O2
400g will produce = (96* 400)/245 = 156.7g of O2
Theoretical yield = 115 g (given)
%yield = (115/156.7)* 100 = 73.39% = 73.40%
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