Calculate how many grams of the first reactant are necessary to completely react with 55.8g of the second reactant.
Na2CO3 + CuCl2 → CuCO3 + 2 NaCl
55.8gCuCl2 x (1molCuCl2/134.45gCuCl2) x (1molNa2CO3/1molCuCl2) x (105.99gNaCO3/molNa2CO3) =44.0gNa2CO3
I Think
Yes, I get 43.988 g which rounds to 44.0 to 3 s.f.
13.8g
To calculate the grams of the first reactant needed to completely react with the second reactant, we need to determine the molar ratio between the two reactants.
1. Start by writing down the balanced chemical equation:
Na2CO3 + CuCl2 → CuCO3 + 2 NaCl
2. Determine the molar masses of the two reactants:
- Na2CO3 (Sodium carbonate) has a molar mass of 105.99 g/mol.
- CuCl2 (Copper(II) chloride) has a molar mass of 134.45 g/mol.
3. Calculate the molar ratio between the two reactants based on the balanced equation:
From the balanced equation, we see that the molar ratio between Na2CO3 and CuCl2 is 1:1. This means that for every 1 mole of CuCl2, we need 1 mole of Na2CO3.
4. Convert the given mass of the second reactant (CuCl2) to moles:
Using the molar mass of CuCl2 (134.45 g/mol), we can convert the mass to moles:
55.8 g CuCl2 * (1 mol CuCl2 / 134.45 g) = 0.4157 mol CuCl2
5. Use the molar ratio to determine the moles of the first reactant (Na2CO3):
Since the molar ratio between Na2CO3 and CuCl2 is 1:1, the moles of Na2CO3 needed would be the same as the moles of CuCl2:
0.4157 mol Na2CO3
6. Convert the moles of Na2CO3 to grams:
Using the molar mass of Na2CO3 (105.99 g/mol), we can convert the moles to grams:
0.4157 mol Na2CO3 * (105.99 g / 1 mol) = 43.96 g Na2CO3
Therefore, you would need 43.96 grams of Na2CO3 to completely react with 55.8 grams of CuCl2.
It's just another stoichiometry problem.
mols CuCl2 = grams/molar mass = approx 0.4 but you need to be more accurate than that.
Then 0.4 mol CuCl2 x (1 mol Na2CO3/1 mol CuCl2) = 0.4 x 1/1 = 0.4 mol Na2CO3. To convert that to grams you have
g = mols Na2CO3 x molar mass Na3CO3.