QuestionHow many moles of $\mathrm{NH}_{3}$ are produced from 4 moles of $\mathrm{H}_{2}$ with excess $\mathrm{N}_{2}$ ? $3 \mathrm{H}_{2}+\mathrm{N}_{2} \rightarrow 2 \mathrm{NH}_{3}$

Studdy Solution

STEP 1

Assumptions1. The reaction is a complete reaction, meaning all reactants are completely converted into products. . The reaction follows the stoichiometry given by the balanced chemical equation $3 \mathrm{H}_{}+\mathrm{}_{} \rightarrow \mathrm{NH}_{3}$ .
3. We have4 moles of $\mathrm{H}_{}$ .
4. $\mathrm{}_{}$ is in excess, meaning there is more than enough $\mathrm{}_{}$ to react with all the $\mathrm{H}_{}$ .

STEP 2

From the balanced chemical equation, we can see that moles of $\mathrm{H}_{2}$ react to form2 moles of $\mathrm{NH}_{}$ .

STEP 3

We can set up a ratio to find out how many moles of $\mathrm{NH}_{3}$ are formed from moles of $\mathrm{H}_{2}$ .
$\frac{2 \, \text{moles of} \, \mathrm{NH}_{3}}{3 \, \text{moles of} \, \mathrm{H}_{2}} = \frac{x \, \text{moles of} \, \mathrm{NH}_{3}}{ \, \text{moles of} \, \mathrm{H}_{2}}$

STEP 4

olving for $x$ will give us the number of moles of $\mathrm{NH}_{3}$ formed. Cross multiply to solve for $x$ .
$x = \frac{2 \, \text{moles of} \, \mathrm{NH}_{3} \times4 \, \text{moles of} \, \mathrm{H}_{2}}{3 \, \text{moles of} \, \mathrm{H}_{2}}$

STEP 5

Calculate the value of $x$ .
$x = \frac{2 \times4}{3} = \frac{8}{3} \approx2.67 \, \text{moles}$ So, if complete reaction occurs with4 moles of $\mathrm{H}_{2}$ and excess $\mathrm{}_{2}$ , approximately2.67 moles of $\mathrm{NH}_{3}$ will be formed.

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