Question

How many grams of Sg is required to produce 83.10 g SF6? S: +24F-->8SF

Answer 1

Answer : The mass of $S_8$ required is 18.238 grams.

Explanation : Given,

Mass of $SF_6$ = 83.10 g

Molar mass of $SF_6$ = 146 g/mole

Molar mass of $S_8$ = 256.52 g/mole

The balanced chemical reaction is,

$S_8+24F_2\rightarrow 8SF_6$

First we have to determine the moles of $SF_6$.

$\text{Moles of }SF_6=\frac{\text{Mass of }SF_6}{\text{Molar mass of }SF_6}=\frac{83.10g}{146g/mole}=0.569moles$

Now we have to determine the moles of $S_8$.

From the balanced chemical reaction we conclude that,

As, 8 moles of $SF_6$ produced from 1 mole of $S_8$

So, 0.569 moles of $SF_6$ produced from $\frac{0.569}{8}=0.0711$ mole of $S_8$

Now we have to determine the mass of $S_8$.

$\text{Mass of }S_8=\text{Moles of }S_8\times \text{Molar mass of }S_8$

$\text{Mass of }S_8=(0.0711mole)\times (256.52g/mole)=18.238g$

Therefore, the mass of $S_8$ required is 18.238 grams.