Question

what mass of Ca(OH)2 will react with 25.0 g of phosphoric acid to form the preservative calcium propionate according to the equation?

Answer 1

Answer: The mass of calcium hydroxide reacted is 28.305 g

Step-by-step explanation:

To calculate the number of moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` .....(1)

Given mass of phosphoric acid = 25 g

Molar mass of phosphoric acid = 98 g/mol

Putting values in equation 1, we get:

`\text{Moles of phosphoric acid}=(25g)/(98g/mol)=0.255mol`

The chemical equation for the reaction of phosphoric acid and calcium hydroxide follows:

`3Ca(OH)_2+2H_3PO_4\rightarrow Ca_3(PO_4)_2+6H_2O`

By Stoichiometry of the reaction:

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

So, 0.255 moles of phosphoric acid will react with =
`(3)/(2)* 0.255=0.3825moles` of calcium hydroxide.

Now, calculating the mass of calcium hydroxide from equation 1, we get:

Molar mass of calcium hydroxide = 74 g/mol

Moles of calcium hydroxide = 0.3825 moles

Putting values in equation 1, we get:

`0.3825mol=\frac{\text{Mass of calcium hydroxide}}{74g/mol}\\\\\text{Mass of calcium hydroxide}=28.305g`

Hence, the mass of calcium hydroxide reacted is 28.305 g