According to the equation below, how many moles of Ca(OH)2 are required to react with 1.36 mol H3PO4 to produce Ca3(PO4)2? 3Ca(OH)2+2H3PO4⟶Ca3(PO4)2+6H2O

The amount of calcium hydroxide needed to react is 2.04 moles

Explanation:

We are given:

Moles of phosphoric acid = 1.36 moles

For the given chemical equation:

By Stoichiometry of the reaction:

2 moles of phosphoric acid reacts with 3 moles of calcium hydroxide

So, 1.36 moles of phosphoric acid will react with = of calcium hydroxide

Hence, the amount of calcium hydroxide needed to react is 2.04 moles

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o