Sodium bromate reacts with calcium
What is the percentage yield if 82.7 g of sodium bromate reacts and 62.55 g of the calcium product is experimentally produced?
NaBrO3 + Cl2
Molar Mass of NaBrO3 = 150.89
= 82.7/150.89 = 0.55
= (0.55 × 62.55)/82.7= 0.4145×100
= 41.45%
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