In a reaction, 39.5 g of chromium(III) oxide reacts with 14.0 g of aluminum to produce chromium and aluminum oxide. If 27.0 g of chromium is produced, what mass of aluminum oxide is produced?

I don't know where to begin. Do I put this into a chemical formula? balance it out?
like,
2 CR2O3 + 3Al --> 4 Cr + 3 AlO2??

In the book it said something about ratios, but I have no clue how to do that. Please help explain this, much appreciated!

i can help you as soon as you tell me your grade

1. Write the equation and balance it.

2. Convert 27.0 g Cr to moles. moles = grams Cr/atomic mass Cr.
3. Using the coefficients in the balanced equation (that's the ratio the book is talking about), convert moles Cr to moles Al. It will look like this.
moles Cr(from step 1) x (y moles Al/z moles Cr) = moles Al.
moles Cr you have from step 1, y moles Al where y is the coefficient in the balanced equation for Al and z is the coefficient in the balanced equation for Cr. Note how the factor (the part in parentheses) cancels Cr but leaves Al. That is how we convert moles of one material in a chemical equation to any other material we wish.
4. Now convert moles Al to grams. g = moles x molar mass.
This procedure is a general one for stoichiometry problems in chemistry.

To solve this problem, we can start by balancing the chemical equation. The balanced equation you wrote is correct:

2 Cr2O3 + 3 Al -> 4 Cr + 3 Al2O3

Now, we need to calculate the amount of aluminum oxide produced. We can use the concept of stoichiometry and the given mass of chromium produced to determine the mass of aluminum oxide.

First, calculate the moles of chromium produced:
molar mass of Cr = 52 g/mol
moles of Cr = mass of Cr / molar mass of Cr
moles of Cr = 27.0 g / 52 g/mol ≈ 0.519 mol

According to the balanced equation, the stoichiometric ratio between chromium and aluminum oxide is 4:3. This means that for every 4 moles of chromium produced, 3 moles of aluminum oxide are produced.

Using this ratio, we can find the amount of aluminum oxide produced:
moles of Al2O3 = (moles of Cr) x (3 moles of Al2O3 / 4 moles of Cr)
moles of Al2O3 = 0.519 mol x (3/4) ≈ 0.389 mol

Finally, calculate the mass of aluminum oxide produced:
molar mass of Al2O3 = 102 g/mol
mass of Al2O3 = moles of Al2O3 x molar mass of Al2O3
mass of Al2O3 = 0.389 mol x 102 g/mol ≈ 39.7 g

Therefore, approximately 39.7 g of aluminum oxide is produced.

To solve this question, you need to use stoichiometry, which is the calculation of the quantities of reactants and products in a chemical reaction. Here's how to approach it step by step:

1. Write out the balanced chemical equation for the reaction as you did:
2 Cr2O3 + 3 Al --> 4 Cr + 3 Al2O3

2. Calculate the molar masses:
- Molar mass of Cr2O3 (chromium(III) oxide) = 2(52.00 g/mol) + 3(16.00 g/mol) = 152.00 g/mol
- Molar mass of Al (aluminum) = 26.98 g/mol
- Molar mass of Cr (chromium) = 52.00 g/mol
- Molar mass of Al2O3 (aluminum oxide) = 2(26.98 g/mol) + 3(16.00 g/mol) = 101.96 g/mol

3. Convert the given masses of reactants to moles:
- Moles of Cr2O3 = 39.5 g / 152.00 g/mol = 0.259 moles
- Moles of Al = 14.0 g / 26.98 g/mol = 0.519 moles

4. Determine the limiting reactant:
The limiting reactant is the one that is entirely consumed in the reaction and limits the amount of product formed. To find the limiting reactant, compare the ratios of the stoichiometric coefficients from the balanced equation to the quantities in moles.

From the balanced equation, the stoichiometric ratio is 2:3 for Cr2O3 to Al. So, using the mole quantities calculated above, you can see that there is an excess of Al.

Therefore, Al is the limiting reactant, and Cr2O3 is in excess.

5. Determine the moles of Cr produced:
From the balanced equation, the stoichiometric ratio is 2:4 for Cr2O3 to Cr. So, since you have 0.259 moles of Cr2O3, you can calculate the moles of Cr produced as follows:

Moles of Cr = (4 moles Cr / 2 moles Cr2O3) * 0.259 moles = 0.518 moles

6. Calculate the mass of Al2O3 produced:
From the balanced equation, the stoichiometric ratio is 3:3 for Al to Al2O3. So, since you have 0.519 moles of Al, you can calculate the moles of Al2O3 produced as follows:

Moles of Al2O3 = (3 moles Al2O3 / 3 moles Al) * 0.519 moles = 0.519 moles

Finally, to find the mass of Al2O3 produced, multiply the moles of Al2O3 by its molar mass:

Mass of Al2O3 = 0.519 moles * 101.96 g/mol = 52.97 g

Therefore, the mass of aluminum oxide produced is approximately 52.97 g.