Question

For the reaction:

4Cr(s)+3O2(g)→2Cr2O3(s)
calculate how many grams of the product form when 20.4 g of O2 completely reacts.
Assume that there is more than enough of the other reactant.

Answer 1

Step-by-step explanation:

To solve this problem, we need to first determine which reactant is limiting and which is in excess. We can do this by using stoichiometry and the given amounts of the reactants.

From the balanced chemical equation, we know that 4 moles of Cr react with 3 moles of O2 to produce 2 moles of Cr2O3. Therefore, we can set up a ratio:

4 mol Cr : 3 mol O2 : 2 mol Cr2O3

To find the amount of Cr2O3 produced from 20.4 g of O2, we need to first convert the mass of O2 to moles:

20.4 g O2 x (1 mol O2/32.00 g O2) = 0.6375 mol O2

Using the mole ratio from the balanced equation, we can determine the number of moles of Cr2O3 produced:

0.6375 mol O2 x (2 mol Cr2O3/3 mol O2) = 0.425 mol Cr2O3

Finally, we can convert moles of Cr2O3 to grams using the molar mass of Cr2O3:

0.425 mol Cr2O3 x (151.99 g Cr2O3/1 mol Cr2O3) = 64.68 g Cr2O3

Therefore, 64.68 grams of Cr2O3 are produced when 20.4 g of O2 completely reacts.