Question

PRILOGAHEThe HandUse the balanced equation to solve the problem.N₂ + 3H₂ -----> 2NH323.0g NH3 are made.How many liters of H₂ gas reacted at STP?L

Answer 1

45.47 liters of H₂ gas reacted at STP.

Step-by-step 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:

`Moles=\frac{Mass}{Molar\text{ }mass}`

Putting the mass of NH₃ = 23.0 g and the molar mass of NH₃ = 17.031 g/mol, we have

`\begin{gathered} Moles=\frac{23.0g}{17.031g\text{/}mol} \\ \\ Moles=1.350478539\text{ }mol \end{gathered}`

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.

`\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}*3mol\text{ }H₂\text{ }gas \\ \\ x=2.03\text{ }mol\text{ }H_2\text{ }gas \end{gathered}`

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

`\begin{gathered} x=\frac{2.03mol}{1.0\text{ }mol}*22.4\text{ }L \\ \\ x=45.47\text{ }L \end{gathered}`

Hence, the liters of H₂ that reacted to form 23.0 g of NH₃ is 45.47 L