We’re being asked to calculate the percent yield of CH3Br in the reaction of 5.00 g HBr with excess CH4O. The balanced chemical equation is:
CH4O + HBr → CH3Br + H2O
Recall that percent yield is given by:
We know the reaction produced 12.23 g CH3Br; this is the actual yield of the reaction. To calculate the theoretical yield, we need to do the following:
Mass of CH4O (molar mass of CH4O) → Moles of CH4O (mole-to-mole comparison) → Moles of CH3Br (molar mass of CH3Br) → Mass of CH3Br
Since HBr is in excess, we can simply ignore it in our calculations.
What is the percent yield for CH3Br if 12.23 g is produced when 5.00 g of CH4O reacts with excess HBr? CH4O + HBr → CH3Br + H2O
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