Question

A sample of calcium oxide whose mass was 12g treated with 46g of water to prepare Calcium

hydroxide?

a) Which compound was limiting reagent?

b) Calculate the amount of Ca(OH)2 formed?

c) Calculate the excess amount of Excess reagent​

Answer 1

Answer:

Explanation:

CaO+H  

2

​  

O→Ca(OH)  

2

​  

 

One mole of CaO will give one mole of calcium hydroxide.

The molar mass of calcium hydroxide is 40.1+2(16+1)=74.1g/mol.  

370 grams of calcium hydroxide corresponds to  

74.1g/mol

370g

​  

=5.0  moles.

Hence, 5.0 moles of CaO are required  to react with an excess of water to form 370 grams of calcium hydroxide

Answer 2

Explanation:

CaO+H

2

O→Ca(OH)

2

One mole of CaO will give one mole of calcium hydroxide.

The molar mass of calcium hydroxide is 40.1+2(16+1)=74.1g/mol.

370 grams of calcium hydroxide corresponds to

74.1g/mol

370g

=5.0 moles.

Hence, 5.0 moles of CaO are required to react with an excess of water to form 370 grams of calcium hydroxide