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

Answer:

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.

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 = $\dfrac{mass}{molar \ mass}$

∴

number of moles of Cr2S3 = $\dfrac{421}{200.19}$

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 $Cr_2O_3_{(s)}$