Na2SO4(s) + 4 C(s) = 1 Na2S(s) + 4 CO(g) (b) If you start with 92.5 g of sodium sulfate and 21.7 g carbon, calculate the maximum mass of sodium sulfide that could be produced.

Respuesta :

Answer:

The maximum mass of sodium sulfide that could be produced is 35.26 grams.

Explanation:

[tex]Na_2SO_4(s) + 4C(s)\rightarrow 1 Na_2S(s) + 4 CO(g)[/tex]

Moles of sodium sulfate = [tex]\frac{92.5 g}{142 g/mol}=0.6514 mol[/tex]

Moles of carbon = [tex]\frac{21.7 g}{12 g/mol}=1.8083 mol[/tex]

According to reaction, 1 mole of sodium sulfate reacts with 4 moles of carbon.

Then 0.6514 moles of sodium sulfate will react with:

[tex]\frac{4}{1}\times 0.6514 mol=2.6056 mol[/tex] of carbon.

2.6056 moles of carbon > 1.8083 mol of carbon (given)

As we can see that moles of sodium sulfate are in excess ,so, the amount of sodium sulfide will depend upon moles of carbon.

According to reaction, 4 mole of carbon gives with 1 moles of sodium sulfide.

Then 1.8083 moles of carbon will give :

[tex]\frac{1}{4}\times 1.8083 mol=0.4521 mol[/tex] of sodium sulfide.

Mass of 0.4521 moles of sodium sulfide:

0.4521 mol × 78 g/mol = 35.26 g

The maximum mass of sodium sulfide that could be produced is 35.26 grams.