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The following balanced equation shows the decomposition of ammonia (NH_(3)) into nitrogen (N_(2)) and hydrogen (H_(2)) 2NH_(3)arrow N_(2)+3H_(2) Aquantity of NH_(3) decomposes to produce 0.0351 mol N_(2) How many moles of H_(2) are produced? 0.0351molH_(2) 0.0117molH_(2) 0.105molH_(2) 0.0702molH_(2)

Question

The following balanced equation shows the decomposition of ammonia (NH_(3)) into nitrogen (N_(2))
and
hydrogen (H_(2))
2NH_(3)arrow N_(2)+3H_(2)
Aquantity of NH_(3) decomposes to produce 0.0351 mol N_(2) How many moles of H_(2) are produced?
0.0351molH_(2)
0.0117molH_(2)
0.105molH_(2)
0.0702molH_(2)

The following balanced equation shows the decomposition of ammonia (NH_(3)) into nitrogen (N_(2)) and hydrogen (H_(2)) 2NH_(3)arrow N_(2)+3H_(2) Aquantity of NH_(3) decomposes to produce 0.0351 mol N_(2) How many moles of H_(2) are produced? 0.0351molH_(2) 0.0117molH_(2) 0.105molH_(2) 0.0702molH_(2)

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MeredithMaster · Tutor for 5 years

Answer

### \( 0.105 \mathrm{~mol} \mathrm{H}_{2} \)

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## Step 1: Identify Mole Ratio<br />The balanced equation for the decomposition of ammonia (\( \mathrm{NH}_3 \)) is given by:<br />\[ 2 \mathrm{NH}_3 \rightarrow \mathrm{N}_2 + 3 \mathrm{H}_2 \]<br />From the equation, we see that 1 mole of \( \mathrm{N}_2 \) is produced for every 3 moles of \( \mathrm{H}_2 \).<br /><br />## Step 2: Use Mole Ratio<br />Given \( 0.0351 \) mol of \( \mathrm{N}_2 \) is produced, we can find the moles of \( \mathrm{H}_2 \) using the mole ratio \( \frac{3 \text{ mol } \mathrm{H}_2}{1 \text{ mol } \mathrm{N}_2} \):<br />\[ \text{Moles of } \mathrm{H}_2 = 0.0351 \text{ mol } \mathrm{N}_2 \times \frac{3 \text{ mol } \mathrm{H}_2}{1 \text{ mol } \mathrm{N}_2} \]<br />\[ = 0.1053 \text{ mol } \mathrm{H}_2 \]<br /><br />## Step 3: Select the Closest Answer<br />The closest option to \( 0.1053 \) mol is \( 0.105 \) mol \( \mathrm{H}_2 \).<br /><br />#
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