Question

Consider the following unbalanced reaction:

C_(s) + S_8(s) → CS_2(l)

How many grams of carbon disulfide will be produced if 6.00 g of C and 10.3 g of S₈ are allowed to react completely?

Answer 1

The balanced equation is

4 C (s) + S₈ (s) → 4 CS₂ (l)

Number of moles of C = 6.00 g / 12.0107 g/mol = 0.500 mole

Number of moles of S8 = 10.3 g / 256.5240 g/mol = 0.0402 mole

From the balanced equation we can say that

4 mole of C requires 1 mole of S8 so

0.500 mole of C will require

= 0.500 mole of C *(1 mole of S8 / 4 mole of C)

= 0.125 mole of S8

But we have 0.0402 mole of S8 which is in short so S8 is limiting reactant

From the balanced equation we can say that

1 mole of S8 produces 4 mole of CS2 so

0.0402 mole of S8 will produce

= 0.0402 mole of S8 *(4 mole of CS2 / 1 mole of S8)

= 0.161 mole of CS2

mass of 1 mole of CS2 = 76.139 g so

the mass of 0.161 mole of CS2 = 12.3 g

Therefore, the mass of CS2 produced would be 12.3 g