if a mixture of 90 g of hydrogen sulfide and 70.5 g of chromium oxide are allowed to raeact what mass of water can be formed

24.84 g

Explanation:

The balanced reaction equation is;

Cr2O3(s) + 3H2S(g) ⟶Cr2S3(s) + 3H2O(l)

We must first determine the limiting reactant

For chromium III oxide;

Amount of chromium III oxide = mass/molar mass = 70.5g/ 151.99 g/mol = 0.46 moles

If 1 mole of chromium III oxide yields 3 moles of water

0.46 moles of chromium III oxide yields 0.46 × 3 = 1.38 moles of water

For hydrogen sulphide

Amount of hydrogen sulphide = mass/molar mass = 90g/ 34 gmol-1 = 2.64 moles

If 3 moles of H2S yields 3 moles of water

2.64 moles of H2S yields 2.64 × 3/3 = 2.64 moles of water

Hence chromium III oxide is the limiting reactant.

Mass of water produced= 1.38 moles × 18gmol-1= 24.84 g

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