Question

10.4 g of As reacts with 11.8g of S to produce 14.2 g of AssS3. Find the limiting reactant, theoretical yield and percent yield for the reaction. The balance chemical equation is:

Answer 1

$n = \frac{m}{M}$

n(As) $= \frac{10.4}{74.9} = 0.1388 \; mol$

n(S) $= \frac{11.8}{32.06} = 0.3680 \;mol$

Theoretical proportion of As : S = 2 : 3 = 1 : 1.5

Real proportion of As : S = 0.138 : 0.368 = 1 : 2.6

So, As is the limitimg reactant.

n(As₂S₃) $= \frac{1}{2}n(As) = \frac{0.1388}{2} = 0.0694 \;mol$

M(As₂S₃) = 246.04 g/mol

m(As₂S₃) $= 0.0694 \times 246.04 = 17.07\; g$ (theoretical yield).

Proportion:

17.07 – 100 %

14.2 – x

$x = \frac{14.2 \times 100}{17.07} = 83.18$ % (percent yield for the reaction)