Respuesta :
Answer: The theoretical yield of sodium sulfate is 1.30 grams.
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
Given mass of rhodium(III) hydroxide = 0.730 g
Molar mass of rhodium(III) hydroxide = 120 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of rhodium(III) hydroxide}=\frac{0.730g}{120g/mol}=0.0061mol[/tex]
The given chemical equation follows:
[tex]Rh_2(SO_4)_3(aq.)+6NaOH(aq.)\rightarrow 2Rh(OH)_3(s)+3Na_2SO_4(aq.)[/tex]
By Stoichiometry of the reaction:
When 2 moles of rhodium(III) hydroxide are produced, then 3 moles of sodium sulfate is also produced
So, when 0.0061 moles of rhodium(III) hydroxide are produced, then = [tex]\frac{3}{2}\times 0.0061=0.00915mol[/tex] of sodium sulfate is also produced
Now, calculating the mass of sodium sulfate from equation 1, we get:
Molar mass of sodium sulfate = 142 g/mol
Moles of sodium sulfate = 0.00915 moles
Putting values in equation 1, we get:
[tex]0.00915mol=\frac{\text{Mass of sodium sulfate}}{142g/mol}\\\\\text{Mass of sodium sulfate}=(0.00915mol\times 142g/mol)=1.30g[/tex]
Hence, the theoretical yield of sodium sulfate is 1.30 grams.
