Calcium oxide reacts with water in a combination reaction to produce calcium hydroxide: CaO (s) H 2O (l) Ca(OH) 2 (s) In a particular experiment, a 2.00-g sample of CaO is reacted with excess water and 2.14 g of Ca(OH) 2 is recovered. What is the percent yield in this experiment

Find the moles of CaO

divide mass (2.0g) by the RFM which is 56 (Ca is 40 add that to O which is 16 making 56) this gives 0.0356 moles.

Find the theoretical mass by multiplying the moles of CaO (which is 0.0356 as there are no balancing number making the ratio 1:1) by the RFM of Ca(OH)2 which is 74 (40+16+16+1+1)

74 (Ca(OH)2 RFM) x 0.0357 (CaO moles) = 2.6g which is the theoretical mass of Ca(OH)2

Find percentage yield by dividing the actual mass of Ca(OH)2 by the theoretical and then x100 this Should give you 82.3%

It can be classified based on its