To determine the mass of \( \text{H}_2 \) that must have reacted to form \( 0.575 \) grams of \( \text{NH}_3 \), we will follow a step-by-step approach.
1. Calculate the moles of \( \text{NH}_3 \) formed:
We start by using the molar mass of \( \text{NH}_3 \):
[tex]\[
\text{Molar mass of } \text{NH}_3 = 17.03 \text{ g/mol}
\][/tex]
Using the given mass of \( \text{NH}_3 \):
[tex]\[
\text{Mass of } \text{NH}_3 \text{ formed} = 0.575 \text{ g}
\][/tex]
We calculate the moles of \( \text{NH}_3 \) as follows:
[tex]\[
\text{Moles of } \text{NH}_3 = \frac{\text{Mass of } \text{NH}_3}{\text{Molar mass of } \text{NH}_3} = \frac{0.575 \text{ g}}{17.03 \text{ g/mol}} \approx 0.0338 \text{ mol}
\][/tex]
2. Use stoichiometry to find the moles of \( \text{H}_2 \) needed:
From the balanced chemical equation:
[tex]\[
3 \text{ H}_2 + \text{ N}_2 \rightarrow 2 \text{ NH}_3
\][/tex]
This tells us that 3 moles of \( \text{H}_2 \) produce 2 moles of \( \text{NH}_3 \).
Thus, the number of moles of \( \text{H}_2 \) needed can be found using the ratio:
[tex]\[
\text{Moles of } \text{H}_2 = \left( \frac{3}{2} \right) \times \text{Moles of } \text{NH}_3 = \left( \frac{3}{2} \right) \times 0.0338 \text{ mol} \approx 0.0506 \text{ mol}
\][/tex]
3. Calculate the mass of \( \text{H}_2 \) needed:
Using the molar mass of \( \text{H}_2 \):
[tex]\[
\text{Molar mass of } \text{H}_2 = 2.0158 \text{ g/mol}
\][/tex]
We can now find the mass of \( \text{H}_2 \) needed:
[tex]\[
\text{Mass of } \text{H}_2 = \text{Moles of } \text{H}_2 \times \text{Molar mass of } \text{H}_2 = 0.0506 \text{ mol} \times 2.0158 \text{ g/mol} \approx 0.102 \text{ g}
\][/tex]
4. Determine the mass of \( \text{H}_2 \) to the correct number of significant figures:
Given the values provided and the calculation, the mass of \( \text{H}_2 \) to three significant figures would be:
[tex]\[
\boxed{0.102 \text{ grams}}
\][/tex]
Thus, the mass of [tex]\( \text{H}_2 \)[/tex] that must have reacted is [tex]\( 0.102 \)[/tex] grams.
