Question

Determine the mass of As2S3 produced if 10.0 g AsCl3 and 13.4 g H2S are reacted according to the following reaction:

2AsCl3 + 3H2S à As2S3 + 6HCl

Answer 1

2AsCl₃ + 3H₂S → As₂S₃ + 6HCl

M(AsCl₃) = 181.28 g/mol

n = m/M

n(AsCl₃) $= \frac{10.0}{181.28} = 0.055\;mol$

M(H₂S) = 34.1 g/mol

n(H₂S) $= \frac{13.4}{34.1} = 0.382\; mol$

For every 2 moles of AsCl₃ we need 3 moles of H₂S.

For 0.055 moles of AsCl₃ we need only 0.0825 moles of H₂S.

AsCl₃ is the limiting reactant.

$\frac{0.055}{2} = 0.0275$ moles of As₂S₃ will be produced from 0.055 moles of AsCl_3.

$m = n\times M$

M(As₂S₃) = 246.02 g/mol

m(As₂S₃) $= 0.0275\times 246.02 = 6.76\;g$

Answer: 6.76 g