We’re being asked to calculate the percent yield of Ca(OH)2 in the reaction of 1.50 g CaO with excess H2O. The balanced chemical equation is:
CaO(s) + H2O(l) → Ca(OH)2(s)
Recall that percent yield is given by:
We know the reaction produced 1.48 g Ca(OH)2; this is the actual yield of the reaction. To calculate the theoretical yield, we need to do the following:
Mass of CaO (molar mass of CaO) → Moles of CaO (mole-to-mole comparison) → Moles of Ca(OH)2 (molar mass of Ca(OH)2) → Mass of Ca(OH)2
Since H2O is in excess, we can simply ignore it in our calculations.
Calcium oxide reacts with water in a combination reaction to produce calcium hydroxide:
CaO (s) + H2O (l) → Ca(OH)2 (s)
In a particular experiment, a 1.50 g sample of CaO is reacted with 1.50 kg of water and 1.48 g of Ca(OH)2 is recovered. What is the percent yield in this experiment?
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