Answer: 3.024 g grams of hydrogen are needed to convert 76 grams of chromium(III) oxide, ![Cr_{2}O_{3}](https://tex.z-dn.net/?f=Cr_%7B2%7DO_%7B3%7D)
Explanation:
The reaction equation for given reaction is as follows.
![Cr_{2}O_{3} + 3H_{2} \rightarrow 2Cr + 3H_{2}O](https://tex.z-dn.net/?f=Cr_%7B2%7DO_%7B3%7D%20%2B%203H_%7B2%7D%20%5Crightarrow%202Cr%20%2B%203H_%7B2%7DO)
Here, 1 mole of
reacts with 3 moles of
.
As mass of chromium (III) oxide is given as 76 g and molar mass of chromium (III) oxide
is 152 g/mol.
Number of moles is the mass of substance divided by its molar mass. So, moles of
is calculated as follows.
![No. of moles = \frac{mass}{molar mass}\\= \frac{76 g}{152 g/mol}\\= 0.5 mol](https://tex.z-dn.net/?f=No.%20of%20moles%20%3D%20%5Cfrac%7Bmass%7D%7Bmolar%20mass%7D%5C%5C%3D%20%5Cfrac%7B76%20g%7D%7B152%20g%2Fmol%7D%5C%5C%3D%200.5%20mol)
Now, moles of
.given by 0.5 mol of
is calculated as follows.
![0.5 mol Cr_{2}O_{3} \times \frac{3 mol H_{2}}{1 mol Cr_{2}O_{3}}\\= 1.5 mol H_{2}](https://tex.z-dn.net/?f=0.5%20mol%20Cr_%7B2%7DO_%7B3%7D%20%5Ctimes%20%5Cfrac%7B3%20mol%20H_%7B2%7D%7D%7B1%20mol%20Cr_%7B2%7DO_%7B3%7D%7D%5C%5C%3D%201.5%20mol%20H_%7B2%7D)
As molar mass of
is 2.016 g/mol. Therefore, mass of
is calculated as follows.
![No. of moles = \frac{mass}{molar mass}\\1.5 mol = \frac{mass}{2.016 g/mol}\\mass = 3.024 g](https://tex.z-dn.net/?f=No.%20of%20moles%20%3D%20%5Cfrac%7Bmass%7D%7Bmolar%20mass%7D%5C%5C1.5%20mol%20%3D%20%5Cfrac%7Bmass%7D%7B2.016%20g%2Fmol%7D%5C%5Cmass%20%3D%203.024%20g)
Thus, we can conclude that 3.024 g grams of hydrogen are needed to convert 76 grams of chromium(III) oxide,
.