For the reaction shown, find the limiting reactant for each of the following initial amounts of reactants.
4Cr(s)+3O2(g)→2Cr2O3(s)
1.) 1 molCr; 1 molO2
2.) 4 molCr; 2.5 molO2
3.) 12 molCr; 10 molO2
4.) 14.8 molCr; 10.3 molO2
Express your answer as a chemical formula.
I'm stuck on how you write it as a chemical formula, pleaseee help! Thanks!
To find the limiting reactant, you need to determine the moles of each reactant and then compare their ratios to the stoichiometric coefficients in the balanced equation.
1.) 1 mol Cr; 1 mol O2
In this case, the mole ratio of Cr to O2 is 1:1, which matches the stoichiometric coefficient ratio of 4:3. Since the ratios are the same, neither reactant is limiting.
2.) 4 mol Cr; 2.5 mol O2
By using the stoichiometric coefficients, you can calculate the expected moles of Cr2O3 formed from each reactant.
For Cr: 4 mol Cr × (2 mol Cr2O3 / 4 mol Cr) = 2 mol Cr2O3
For O2: 2.5 mol O2 × (2 mol Cr2O3 / 3 mol O2) = 1.67 mol Cr2O3
Comparing the moles of Cr2O3 formed from each reactant, you can see that 2 mol Cr2O3 (from Cr) is less than 1.67 mol Cr2O3 (from O2). Therefore, Cr is the limiting reactant.
3.) 12 mol Cr; 10 mol O2
Using the same calculation method as above, you can determine the moles of Cr2O3 formed from each reactant.
For Cr: 12 mol Cr × (2 mol Cr2O3 / 4 mol Cr) = 6 mol Cr2O3
For O2: 10 mol O2 × (2 mol Cr2O3 / 3 mol O2) = 6.67 mol Cr2O3
Comparing the moles of Cr2O3 formed, you can see that 6 mol Cr2O3 (from Cr) is less than 6.67 mol Cr2O3 (from O2). Therefore, Cr is again the limiting reactant.
4.) 14.8 mol Cr; 10.3 mol O2
By calculating the moles of Cr2O3 formed:
For Cr: 14.8 mol Cr × (2 mol Cr2O3 / 4 mol Cr) = 7.4 mol Cr2O3
For O2: 10.3 mol O2 × (2 mol Cr2O3 / 3 mol O2) = 6.87 mol Cr2O3
Comparing the moles of Cr2O3 formed, you can see that 7.4 mol Cr2O3 (from Cr) is more than 6.87 mol Cr2O3 (from O2). Therefore, O2 is the limiting reactant.
Thus, the limiting reactant for each of the given initial amounts is:
1.) Neither Cr nor O2 is the limiting reactant.
2.) Cr is the limiting reactant.
3.) Cr is the limiting reactant.
4.) O2 is the limiting reactant.
To determine the limiting reactant, we need to compare the moles of each reactant with the stoichiometric ratio of the balanced chemical equation.
Step 1: Write down the balanced chemical equation:
4Cr(s) + 3O2(g) → 2Cr2O3(s)
Step 2: Convert the given moles of each reactant, one at a time, to moles of the product.
1.) 1 mol Cr and 1 mol O2:
Using the stoichiometric ratio of the balanced equation, we can see that 4 moles of Cr react with 3 moles of O2 to produce 2 moles of Cr2O3. Therefore, the limiting reactant can be determined by comparing the number of moles of product we can obtain if we use each reactant.
Moles of Cr2O3 from 1 mol of Cr = (1 mol Cr) x (2 mol Cr2O3 / 4 mol Cr) = 0.5 mol Cr2O3
Moles of Cr2O3 from 1 mol of O2 = (1 mol O2) x (2 mol Cr2O3 / 3 mol O2) = 0.67 mol Cr2O3
Since the yield of Cr2O3 is lower when using 1 mol of Cr (0.5 mol) compared to 1 mol of O2 (0.67 mol), the limiting reactant is Cr.
To express the answer as a chemical formula, we can simply write "Cr" since it represents the limiting reactant.
You can follow the same steps to find the limiting reactant for the other given scenarios (2, 3, and 4) by comparing the moles of product obtained from each reactant.
I've never done this but I would guess it would be Cr4O3; i.e., for every 4 mols Cr you use 3 mol O2 for #1 with Cr as the limiting reagent.
For #2, O2 is the LR.
2.5 O2 x 4 mol Cr/3 mol O2 = 3.33 so the formula is Cr3.33O2.5. Convert to whole numbers and that is Cr10O7.5 then double that to get rid of 1/2 which makes it Cr20O15. That is divisible by 5 for Cr4O3. ta dah!
I don't know if this is what you want or not.
#3 doesn't work out to be the same as 1 and 2.