How would I find the limiting reagent in the chemical equation...CuSO4*5H2O+2K2C2O4*H2O-->K2[Cu(C2O4)2]*2H2O+K2SO4+5H2O
I got that K2C2O4 was the limiting reagent but I don't know if that is correct.
Here is a link that gives a worked example of a limiting reagent problem. Print this for your use when you need it; the procedure will work all limiting reagent problems.
http://www.jiskha.com/science/chemistry/limiting_reagent_problem.html
To determine the limiting reagent in a chemical equation, you need to compare the stoichiometry of the reactants. The reactant that is completely consumed first and restricts the amount of product formed is the limiting reagent. Here's how to determine the limiting reagent for the given equation:
1. Write down the balanced chemical equation:
CuSO4*5H2O + 2K2C2O4*H2O → K2[Cu(C2O4)2]*2H2O + K2SO4 + 5H2O
2. Convert the amounts of each reactant to moles. Use the molecular weight of each compound to calculate the moles. For example, CuSO4*5H2O has a molar mass of (63.5 + 32.1 + (16.0 * 4) + (1.0 * 2) + (1.0 * 5 * 2)) g/mol.
3. Compare the mole ratios of the reactants in the balanced equation. The coefficients of the reactants represent the mole ratios.
CuSO4*5H2O : K2C2O4*H2O = 1 : 2
CuSO4*5H2O : K2[Cu(C2O4)2]*2H2O = 1 : 1
4. Calculate the moles of product that can be formed from each reactant. Use the mole ratio from the balanced equation to convert the moles of reactant to moles of product.
Moles of K2C2O4*H2O → Moles of K2[Cu(C2O4)2]*2H2O
Moles of CuSO4*5H2O → Moles of K2[Cu(C2O4)2]*2H2O
5. Compare the moles of product that can be formed from each reactant. The reactant that produces the least amount of product is the limiting reagent.
By comparing the two calculations, if the moles of K2C2O4*H2O gas formation are less than the moles of CuSO4*5H2O gas formation, then K2C2O4*H2O is the limiting reagent.
Ultimately, the result depends on the actual amounts or moles of each reactant used in the reaction.
To find the limiting reagent in a chemical equation, you need to compare the moles or the number of molecules of each reactant to determine which one will run out first and limit the amount of product that can be formed.
In this case, you have the chemical equation: CuSO4*5H2O + 2K2C2O4*H2O → K2[Cu(C2O4)2]*2H2O + K2SO4 + 5H2O
To find the limiting reagent, follow these steps:
1. Determine the molar masses of each reactant:
- CuSO4*5H2O: (63.55 + 32.07 + 4 * 16.00) + (5 * (2 * 1.01 + 16.00)) = 249.70 g/mol
- K2C2O4*H2O: (2 * (39.10) + 2 * (12.01 + 2 * 16.00)) + (1 * (2 * 1.01 + 16.00)) = 214.24 g/mol
2. Convert the given amounts of reactants into moles. Let's assume you have 10 grams of CuSO4*5H2O and 15 grams of K2C2O4*H2O.
- Moles of CuSO4*5H2O = 10 g / 249.70 g/mol ≈ 0.040 mol
- Moles of K2C2O4*H2O = 15 g / 214.24 g/mol ≈ 0.070 mol
3. Calculate the stoichiometric ratio of the reactants based on the coefficients in the balanced equation. In this case, the ratio is 1:2 (CuSO4*5H2O : K2C2O4*H2O).
4. Compare the moles of the reactants to the stoichiometric ratio. The reactant with the smaller amount of moles is the limiting reagent.
- CuSO4*5H2O: 0.040 mol
- K2C2O4*H2O: 0.070 mol
Since the stoichiometric ratio is 1:2, for every 0.040 mol of CuSO4*5H2O, you would need 0.080 mol of K2C2O4*H2O. However, you have only 0.070 mol of K2C2O4*H2O.
Therefore, K2C2O4*H2O is the limiting reagent in this reaction.