Question

For the following reaction, 11.3 grams of sulfur are allowed to react with 23.3 grams of carbon monoxide. What amount of the excess reagent remains after the reaction is complete?

Answer 1

Solution:

Given the chemical reaction:

$$S + 2CO → SO_2 + 2C$$

m(sulfur) = 11.3 g

m(carbon monoxide) = 23.3 g

Mr(sulfur) = 32 $(\frac{g}{mole})$

Mr(carbon monoxide) = 28 $(\frac{g}{mole})$

1) We should find the number of moles in order to find limiting reagents:

$n_1=\frac{m_1}{Mr}=\frac{11.3}{32}=0.353\;moles\\n_2=\frac{m_2}{Mr}=0.832\;moles$

but in the chemical reaction, the ratio of coefficients is not 1:1. So, that we should make it 1:1 ratio:

$n_1=0.353\;moles$

$n_2=\dfrac{0.832}{2}=0.416\;moles$

So, in this case, the limiting reactant is sulfur, since it has a smaller mole than moles of carbon monoxide.

2) We should subtract $n_2$ from $n_1$, in order to find the amount of excess reagent:

$n_2-n_1=0.416-0.353=0.063\;moles$

$m_{excess}=n*Mr=0.063*23.3=1.46\;gram$

Answer:

After the reaction is complete, 0.063 moles or 1.46 grams of carbon monoxide has remained.