Respuesta :
Theoretical yield is the ideal number, so we'll assume it all reacted without issue.
[tex] \frac{10.0gH _{2} }{1} [/tex]×[tex] \frac{1molH_{2} }{2.016gH _{2} } [/tex]×[tex] \frac{2molNH _{3} }{3molH _{2} } [/tex]×[tex] \frac{17.03gNH _{3} }{1molNH _{3} } [/tex]=56.3gNH↓3
[tex] \frac{10.0gH _{2} }{1} [/tex]×[tex] \frac{1molH_{2} }{2.016gH _{2} } [/tex]×[tex] \frac{2molNH _{3} }{3molH _{2} } [/tex]×[tex] \frac{17.03gNH _{3} }{1molNH _{3} } [/tex]=56.3gNH↓3
Answer: The correct answer is Option c.
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 hydrogen gas = 10.0 g
Molar mass of hydrogen gas = 2 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of hydrogen gas}=\frac{10.0g}{2g/mol}=5.0mol[/tex]
The given chemical equation follows:
[tex]N_2+3H_2\rightarrow 2NH_3[/tex]
As, nitrogen gas is present in excess. It is considered as an excess reagent.
Hydrogen gas is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
3 moles of hydrogen gas produces 2 moles of ammonia
So, 5.0 moles of hydrogen gas will produce = [tex]\frac{2}{3}\times 5.0=3.33mol[/tex] of ammonia
Now, calculating the mass of ammonia by using equation 1, we get:
Molar mass of ammonia = 17 g/mol
Moles of ammonia = 3.33 moles
Putting values in equation 1, we get:
[tex]3.33mol=\frac{\text{Mass of ammonia}}{17g/mol}\\\\\text{Mass of ammonia}=(3.33mol\times 17g/mol)=56.3g[/tex]
Hence, the correct answer is Option c.