Respuesta :
Answer
45.47 liters of H₂ gas reacted at STP.
Explanation
Given:
Mass in grams of NH₃ made = 23.0 g
The balanced equation for the reaction is:
N₂ + 3H₂ → 2NH₃
What to find:
The liters of H₂ that reacted to form 23.0 g of NH₃.
Step-by-step solution:
Step 1: Determine the mole ratio of H₂ and NH₃ in the equation
The mole ratio of H₂ and NH₃ in the equation is 3:2
Step 2: Convert 23.0 g of NH₃ made to moles.
Using the mole formula:
[tex]Moles=\frac{Mass}{Molar\text{ }mass}[/tex]Putting the mass of NH₃ = 23.0 g and the molar mass of NH₃ = 17.031 g/mol, we have
[tex]\begin{gathered} Moles=\frac{23.0g}{17.031g\text{/}mol} \\ \\ Moles=1.350478539\text{ }mol \end{gathered}[/tex]Step 3: Use the mole ratio in step 1 and the moles of NH₃ in step 2 to determine the moles of H₂ gas that reacted.
[tex]\begin{gathered} 3mol\text{ }H₂\text{ }gas=2mol\text{ }NH_3 \\ \\ x=1.350478539mol\text{ }NH_3 \\ \\ x=\frac{1.350478539mol\text{ }NH_3}{2mol\text{ }NH_3}\times3mol\text{ }H₂\text{ }gas \\ \\ x=2.03\text{ }mol\text{ }H_2\text{ }gas \end{gathered}[/tex]Step 4: Convert the 2.03 moles of H₂ gas to liters.
At STP, 1.0 mole of H₂ gas = 22.4 L
∴ 2.03 moles H₂ gas = x L
To get x, cross multiply and divide both sides by 1 mole
[tex]\begin{gathered} x=\frac{2.03mol}{1.0\text{ }mol}\times22.4\text{ }L \\ \\ x=45.47\text{ }L \end{gathered}[/tex]Hence, the liters of H₂ that reacted to form 23.0 g of NH₃ is 45.47 L
