Asked by Cristin
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
Na2CO3 + CuCl2 → CuCO3 + 2 NaCl
Answers
Answered by
DrBob222
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.
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.
Answered by
ChemHater
55.8gCuCl2 x (1molCuCl2/134.45gCuCl2) x (1molNa2CO3/1molCuCl2) x (105.99gNaCO3/molNa2CO3) =44.0gNa2CO3
I Think
I Think
Answered by
DrBob222
Yes, I get 43.988 g which rounds to 44.0 to 3 s.f.
Answered by
gaby
13.8g
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