Respuesta :
Answer: The actual yield of hydrogen gas in the following reaction is 4.974 grams.
Explanation:
The chemical equation for the reaction of sodium and hydrofluoric acid follows:
[tex]2Na+2HF\rightarrow 2NaF+H_2[/tex]
As, hydrofluoric acid is present in excess. So, it is considered as an excess reagent and sodium is considered as a limiting reagent because it limits the formation of products.
By Stoichiometry of the reaction:
2 moles of sodium is producing 1 mole of hydrogen gas.
So, 5.75 moles of sodium will produce = [tex]\frac{1}{2}\times 5.75=2.875mol[/tex] of hydrogen gas.
To calculate the actual yield of hydrogen gas, we use the equation:
[tex]\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100[/tex]
% yield of reaction = 86.5 %
Theoretical yield of hydrogen gas = 2.875 moles
Putting values in above equation, we get:
[tex]86.5=\frac{\text{Actual yield of hydrogen gas}}{2.875moles}\times 100\\\\\text{Actual yield of hydrogen gas}=2.487mol[/tex]
Now, to calculate the mass of hydrogen gas, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Moles of hydrogen gas = 2.487 mol
Molar mass of hydrogen gas = 2g
Putting values in above equation, we get:
[tex]2.487mol=\frac{\text{Mass of hydrogen gas}}{2g/mol}\\\\\text{Mass of hydrogen gas}=4.974g[/tex]
Hence, the actual yield of hydrogen gas in the following reaction is 4.974 grams.
