Question

GeF3H is formed from GeH4 and GeF4 in the combination reaction: GeH4 + 3GeF4 -> 4GeF3H. If the reactiln yield is 92.6%, how many moles of GeF4 are needed to produce 8 mol of GeF3H?

Answer 1

Answer is: 6,48 mol of GeF₄ are needed.
Chemical reaction: GeH₄ + 3GeF₄ → 4GeF₃H.
n(GeF₃H) = 8 mol.
n(GeF₄) = ?
from chemical reaction: n(GeF₃H) : n(GeF₄) = 4 : 3.
8 mol : n(GeF₄) = 4 : 3.
4n(GeF₄) = 24 mol.
n(GeF₄) = 6 mol ÷ 0,926.
n(GeF₄) = 6,48 mol.

Answer 2

Answer : The number of moles of $GeF_4$ needed to produce 8 mole of $GeF_3H$ are, 6.48 moles.

Explanation :

The given balanced chemical reaction is:

$GeH_4+3GeF_4\rightarrow 4GeF_3H$

First we have to calculate the moles of $GeF_4$

From the balanced chemical reaction we conclude that,

As, 4 moles of $GeF_3H$ produced from 3 mole of $GeF_4$

So, 8 moles of $GeF_3H$ produced from $\frac{8}{4}\times 3=6$ mole of $GeF_4$

Now we have to calculate the moles of $GeF_4$ when the reaction yield is 92.6 %.

As, 92.6 % yield produced from 6 moles of $GeF_4$

So, 100 % yield produced from $\frac{100\%}{92.6\%}\times 6=6.48$ moles of $GeF_4$

Therefore, the number of moles of $GeF_4$ needed to produce 8 mole of $GeF_3H$ are, 6.48 moles.