Question

When 70. G of li3n(s) (molar mass 35 g/mol) reacts with excess h2(g), 8. 0 g of lih(s) is produced. The percent yield is closest to

Answer 1

25%

Step-by-step explanation:

Li3N(s) + 2 H2(g) ⇄ LiNH2(s) + 2 LiH(s)

70 g Li3N x 1 mol Li3N/35 g = 2 mol Li3N

2 mol Li3N x 2 mol LiH/1 mol Li3N

= 4 mol LiH x 7.05 g LiH/1 mol LiH = 28.20

percent yield = actual yield/theoretical yield x 100

8 g / 28.20 g ≈ 28%

28% is closest to 25% ap classrom

Answer 2

The percent yield of LiH is approximately 28.83%.

To find the percent yield, we first need to determine the theoretical yield of LiH using the given reaction and then compare it with the actual yield.

1. **Determine the moles of Li3N:**

`\[ \text{Moles of Li3N} = \frac{\text{Mass of Li3N}}{\text{Molar Mass of Li3N}} = \frac{70 \, \text{g}}{35 \, \text{g/mol}} = 2.0 \, \text{mol} \]`

2. **Use stoichiometry to find moles of LiH:**

From the balanced equation, 1 mol of Li3N produces 2 mol of LiH.

`\[ \text{Moles of LiH} = 2 * \text{Moles of Li3N} = 2 * 2.0 \, \text{mol} = 4.0 \, \text{mol} \]`

3. **Calculate the theoretical yield of LiH:**

`\[ \text{Theoretical Yield of LiH} = \text{Moles of LiH} * \text{Molar Mass of LiH} \]`

`\[ \text{Theoretical Yield of LiH} = 4.0 \, \text{mol} * 6.939 \, \text{g/mol} = 27.756 \, \text{g} \]`

4. **Calculate the percent yield:**

`\[ \text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} * 100 \]`

`\[ \text{Percent Yield} = \frac{8.0 \, \text{g}}{27.756 \, \text{g}} * 100 \approx 28.83\% \]`

The question probable maybe:

Li3N(s) + 2H2(g) ⇄ LiNH2(s) + 2LiH(s) ΔH° = -192 kJ/molrxn Because pure H2 is a hazardous substance, safer and more cost effective techniques to store it as a solid for shipping purposes have been developed. One such method is the reaction represented above, which occurs at 200°C. When 70. g of Li3N(s) (molar mass 35 g/mol) reacts with excess H2(g), 8.0 g of LiH(s) is produced. The percent yield is closest to