Chromium(III) Oxide Reaction: Calculating Precise Quantities

Step 1 of 2

Here we will have to find out amount of \(\mathrm{Cr}_{2} \mathrm{O}_{3}\) is required to produce 421 g of \(\mathrm{Cr}_{2} \mathrm{O}_{3}\).

Chromium(III) oxide reacts with hydrogen sulfide  \(\left(\mathrm{H}_{2} \mathrm{~S}\right)\) gas to form chromium(III) sulfide and water:

\(\mathrm{Cr}_{2} \mathrm{O}_{3}(\mathrm{~s})+3 \mathrm{H}_{2} \mathrm{~S}(\mathrm{~g}) \rightarrow \mathrm{Cr}_{2} \mathrm{~S}_{3}(\mathrm{~s})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\)

(a)  Moles of \(\mathrm{Cr}_{2} \mathrm{O}_{3}\) are required

Given:

Mass of \(\mathrm{Cr}_{2} \mathrm{~S}_{3}\) = 421g

In the above reaction 1 mole of  \(\mathrm{Cr}_{2} \mathrm{~S}_{3}\) has formed from 1 moles of \(\mathrm{Cr}_{2} \mathrm{O}_{3}\).

Molar mass of  \(\mathrm{Cr}_{2} \mathrm{~S}_{3}\) = 200.21 g/mole

We can use the conversion factor = \(\frac{1 \mathrm{~mol} \ \mathrm{Cr}_{2} \mathrm{~S}_{3}}{200.21 \ \mathrm{gCr}_{2} S_{3}}\)

The mole of \(\mathrm{Cr}_{2} \mathrm{O}_{3}\) required can be calculated as,

421 g of \(\mathrm{Cr}_{2} \mathrm{~S}_{3} \times \frac{1 \mathrm{~mol} \ \mathrm{Cr}_{2} \mathrm{~S}_{3}}{200.21 \mathrm{gCr}_{2} \mathrm{~S}_{3}} \times \frac{1 \mathrm{~mol} \ \mathrm{Cr}_{2} \mathrm{O}_{3}}{1 \mathrm{~mol} \ \mathrm{Cr}_{2} \mathrm{~S}_{3}}=2.10 \mathrm{~mol} \text { of } \mathrm{Cr}_{2} \mathrm{O}_{3}\)

Thus  2.10 mol of \(\mathrm{Cr}_{2} \mathrm{O}_{3}\) is required to produce 421g of \(\mathrm{Cr}_{2} \mathrm{~S}_{3}\).