when solid aluminum is heated with chromium(II) oxide, solid chromium and aluminum oxide are produced. If you are using 125 g of aluminum and 225 g of chromium (II) oxide, how many grams of chromium are produced. If you actually collect 110g of chromium wht is the percent yield?
To calculate the grams of chromium produced, you need to determine the limiting reactant in the reaction. The limiting reactant is the one that is completely consumed and determines the maximum amount of product that can be formed.
First, you need to find the molar masses of aluminum and chromium(II) oxide.
- Aluminum (Al) has a molar mass of 26.98 g/mol.
- Chromium(II) oxide (CrO) has a molar mass of 67.97 g/mol.
Next, determine the number of moles for each reactant:
- Moles of aluminum = mass / molar mass = 125 g / 26.98 g/mol = 4.63 mol
- Moles of chromium(II) oxide = mass / molar mass = 225 g / 67.97 g/mol = 3.31 mol
Based on the balanced chemical equation, the stoichiometric ratio of aluminum to chromium is 4:3.
Since the ratio of moles is not 1:1, it is necessary to find the limiting reactant.
The moles of chromium(II) oxide required to react with all the aluminum is:
4.63 mol Al x (3 mol CrO / 4 mol Al) = 3.47 mol CrO
Comparing this with the available moles of chromium(II) oxide, it is evident that 3.31 mol CrO is the limiting reactant.
Using the stoichiometric ratio from the balanced chemical equation:
3.31 mol CrO x (1 mol Cr / 1 mol CrO) = 3.31 mol Cr
To convert moles to grams, multiply by the molar mass of chromium:
3.31 mol Cr x 52 g/mol = 172.12 g Cr
Therefore, the expected mass of chromium produced is 172.12 g.
To calculate the percent yield, divide the actual mass of chromium collected by the expected mass and multiply by 100:
% yield = (actual mass / expected mass) x 100
% yield = (110 g / 172.12 g) x 100 = 63.92%
Thus, the percent yield of chromium is approximately 63.92%.
To determine the quantity of chromium produced when solid aluminum reacts with chromium(II) oxide, we need to calculate the stoichiometry of the reaction and use it to find the molar ratio between aluminum and chromium.
1. Write the balanced equation for the reaction:
2Al + CrO → Al₂O₃ + 2Cr
According to the balanced equation, 2 moles of aluminum react with 1 mole of chromium(II) oxide to produce 2 moles of chromium. Therefore, the molar ratio between aluminum and chromium is 2:2.
2. Calculate the number of moles of aluminum and chromium(II) oxide:
Given the mass of aluminum is 125 g and the molar mass of aluminum is 26.98 g/mol, we can determine the number of moles of aluminum:
Moles of aluminum = Mass of aluminum / Molar mass of aluminum
Moles of aluminum = 125 g / 26.98 g/mol = 4.63 mol
Given the mass of chromium(II) oxide is 225 g and the molar mass of chromium(II) oxide is 110.99 g/mol, we can determine the number of moles of chromium(II) oxide:
Moles of chromium(II) oxide = Mass of chromium(II) oxide / Molar mass of chromium(II) oxide
Moles of chromium(II) oxide = 225 g / 110.99 g/mol = 2.02 mol
3. Determine the limiting reactant:
The limiting reactant is the reactant that is completely consumed in the reaction and restricts the amount of product that can be formed. To find the limiting reactant, we compare the moles of aluminum and chromium(II) oxide.
Since the molar ratio between aluminum and chromium in the balanced equation is 2:2, this means that when the reaction goes to completion, the moles of aluminum will be equal to the moles of chromium produced.
4. Calculate the theoretical yield of chromium:
The theoretical yield is the maximum amount of product that can be obtained from the limiting reactant. In this case, since the moles of aluminum are equal to the moles of chromium produced, the number of moles of chromium produced will also be 4.63 mol.
To convert moles to grams, we use the molar mass of chromium:
Theoretical mass of chromium = Moles of chromium × Molar mass of chromium
Theoretical mass of chromium = 4.63 mol × 51.996 g/mol = 240.2 g
Therefore, the theoretical yield of chromium is 240.2 g.
5. Calculate the percent yield:
Percent Yield = (Actual yield / Theoretical yield) × 100
Given the actual yield of chromium is 110 g, we plug in the values:
Percent Yield = (110 g / 240.2 g) × 100 ≈ 45.8%
The percent yield in this reaction is approximately 45.8%.