Question

Chromium (III) oxide reacts with hydrogen sulfide gas to form chromium (III) sulfide and water.a)writedown the balanced equation for the reaction.b)To produce 421 g of Cr2S3, how many moles of Cr2O3are required

Answer 1

The balanced chemical equation for the reaction can be computed as:

`\mathbf{Cr_2O_3_((s))+ 3H_2S_((g)) \to Cr_2S_3_((s)) + 3H_2O_((l))}`

2.103 moles of
`Cr_2O_3_((s))` is being required.

Step-by-step explanation:

The balanced chemical equation for the reaction can be computed as:

`\mathbf{Cr_2O_3_((s))+ 3H_2S_((g)) \to Cr_2S_3_((s)) + 3H_2O_((l))}`

Recall that numbers of moles =
`(mass)/(molar \ mass)`

∴

number of moles of Cr2S3 =
`(421)/(200.19)`

number of moles of Cr2S3 = 2.103 moles

It obvious from the balanced equation that:

1 mole of
`Cr_2S_3_((s))` requires 1 mole of
`Cr_2O_3_((s))`

Thus; 2.103 moles of
`Cr_2S_3_((s))` = 2.103 × (1/1) moles of
`Cr_2O_3_((s))`

= 2.103 moles of
`Cr_2O_3_((s))`