Question 2f296

Jul 11, 2015

Max amount of $S {O}_{2}$ formed = 23.4 g
Limiting reagent = Sulphur
Amount of excess reagent remained ($C O$) = 3.625 g

Explanation:

First write the balance chemical equation which is as follows;

$S + C O \to S {O}_{2} + C$

Balancing by trial and error method, we get

$S + 2 C O \to S {O}_{2} + 2 C$

From balance chemical equation, we get 1 moles of $S$ reacts with 2 moles of $C O$ to gives 1 moles of $S {O}_{2}$ and 2 moles of Carbon.

We know that,

1 mole of sulphur = 1cancel("mole") * 32"g"/cancel("mole") = "32 g"

2 moles of carbon monoxide = 2cancel("moles") * 28"g"/cancel("mole") = "56 g"

1 mole of sulfur dioxide = 1cancel("mole") * 64"g"/cancel("mole") = "64 g"#

From the balanced chemical equation, you know that 32 g of sulfur (1 mole) requires 56 g of carbon dioxide (2 moles). i.e.
11.7 g of sulfur requires

$\frac{56}{32} \cdot 11.7 = \text{20.475 g}$ $C O$

But here we have 24.1 g of $C O$, which is more than the required 20.475 g. This implies that the limiting reagent is sulfur and the limiting reagent controls the reaction and gives the product.

Therefore, from balance chemical reaction, that if

32 g (1 mole) of $S$ gives 64 g (1 mole) of $S {O}_{2}$

then

11.7 g of $S$ will give

$\frac{64}{32} \cdot 11.7 = \text{23.4 g}$ $S {O}_{2}$

Finally,

We have $C O$ as excess reagent, we know that 11.7 g of $S$ reacts with 20.475 g of $C O$, so the amount excess reagent remained is given by;

${m}_{C O} = \text{24.1 g" - "20.475 g" = "3.625 g}$

Thanks