Answer: The mass of barium nitrate required is 23.91 g
Explanation:
The number of moles is defined as the ratio of the mass of a substance to its molar mass.
The equation used is:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)
Given mass of sodium phosphate = 10.0 g
Molar mass of sodium phosphate = 163.94 g/mol
Plugging values in equation 1:
[tex]\text{Moles of sodium phosphate}=\frac{10.0g}{163.94g/mol}=0.061 mol[/tex]
The chemical equation for the reaction of sodium phosphate and barium nitrate follows:
[tex]3Ba(NO_3)_2+2Na_3PO_4\rightarrow Ba_3(PO_4)_2+6NaNO_3[/tex]
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 = [tex]\frac{3}{2}\times 0.061=0.0915mol[/tex] of barium nitrate
Molar mass of barium nitrate = 261.337 g/mol
Plugging values in equation 1:
[tex]\text{Mass of barium nitrate}=(0.0915mol\times 261.337g/mol)=23.91g[/tex]
Hence, the mass of barium nitrate required is 23.91 g
