Answer
There are necessary 1.33 moles of $NH_3$ and 3.33 moles of $F_2$ to produce 4.00 moles of $HF$.
Work Step by Step
$NH_3:$
1. Find the mole-mole conversion factors:
According to the coefficients: 6 moles $HF$ = 2 moles $NH_3$
$\frac{ 6 \space moles \space HF}{ 2 \space moles \space NH_3}$ and $\frac{ 2 \space moles \space NH_3}{ 6 \space moles \space HF}$
2. Use the mole-mole conversion factors to calculate the amount of equivalent $NH_3$ in moles:
$ 4.00$ moles $HF \times \frac{ 2 \space moles \space NH_3}{ 6 \space moles \space HF} = 1.33$ moles $NH_3$
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$F_2$:
1. Find the mole-mole conversion factors:
According to the coefficients: 6 moles $HF$ = 5 moles $F_2$
$\frac{ 6 \space moles \space HF}{ 5 \space moles \space F_2}$ and $\frac{ 5 \space moles \space F_2}{ 6 \space moles \space HF}$
2. Use the mole-mole conversion factors to calculate the amount of equivalent $F_2$ in moles:
$ 4.00$ moles $HF \times \frac{ 5 \space moles \space F_2}{ 6 \space moles \space HF} = 3.33$ moles $ F_2$
