Question

If 10.0 grams of sodium phosphate are reacted, what mass of barium nitrate is
required?

Answer 1

Answer: The mass of barium nitrate required is 23.91 g

Step-by-step explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

The equation used is:

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

Given mass of sodium phosphate = 10.0 g

Molar mass of sodium phosphate = 163.94 g/mol

Plugging values in equation 1:

`\text{Moles of sodium phosphate}=(10.0g)/(163.94g/mol)=0.061 mol`

The chemical equation for the reaction of sodium phosphate and barium nitrate follows:

`3Ba(NO_3)_2+2Na_3PO_4\rightarrow Ba_3(PO_4)_2+6NaNO_3`

By the stoichiometry of the reaction:

If 2 moles of sodium phosphate reacts with 3 moles of barium nitrate

So, 0.061 moles of sodium phosphate will react with =
`(3)/(2)* 0.061=0.0915mol` of barium nitrate

Molar mass of barium nitrate = 261.337 g/mol

Plugging values in equation 1:

`\text{Mass of barium nitrate}=(0.0915mol* 261.337g/mol)=23.91g`

Hence, the mass of barium nitrate required is 23.91 g