Question

A chemist uses hot hydrogen gas to convert chromium(III) oxide to

pure chromium. How many grams of hydrogen are needed to

convert 76 grams of chromium(III) oxide, Cr203?

Answer 1

Answer: 3.024 g grams of hydrogen are needed to convert 76 grams of chromium(III) oxide,
`Cr_(2)O_(3)`

Step-by-step explanation:

The reaction equation for given reaction is as follows.

`Cr_(2)O_(3) + 3H_(2) \rightarrow 2Cr + 3H_(2)O`

Here, 1 mole of
`Cr_(2)O_(3)` reacts with 3 moles of
`H_(2)`.

As mass of chromium (III) oxide is given as 76 g and molar mass of chromium (III) oxide
`(Cr_(2)O_(3))` is 152 g/mol.

Number of moles is the mass of substance divided by its molar mass. So, moles of
`Cr_(2)O_(3)` is calculated as follows.

`No. of moles = (mass)/(molar mass)\\= (76 g)/(152 g/mol)\\= 0.5 mol`

Now, moles of
`H_(2)`.given by 0.5 mol of
`Cr_(2)O_(3)` is calculated as follows.

`0.5 mol Cr_(2)O_(3) * (3 mol H_(2))/(1 mol Cr_(2)O_(3))\\= 1.5 mol H_(2)`

As molar mass of
`H_(2)` is 2.016 g/mol. Therefore, mass of
`H_(2)` is calculated as follows.

`No. of moles = (mass)/(molar mass)\\1.5 mol = (mass)/(2.016 g/mol)\\mass = 3.024 g`

Thus, we can conclude that 3.024 g grams of hydrogen are needed to convert 76 grams of chromium(III) oxide,
`Cr_(2)O_(3)`.