Asked by Megan
How many grams of C would be necessary to produce 790. grams of CaCO3 by the sequence of reactions shown below?
Step 1: CaO + 3C → CaC2 + CO
Step 2: CaC2 + N2 → CaCN2 + C
Step 3: CaCN2 + 3H2O → CaCO3 + 2NH3
Step 1: CaO + 3C → CaC2 + CO
Step 2: CaC2 + N2 → CaCN2 + C
Step 3: CaCN2 + 3H2O → CaCO3 + 2NH3
Answers
Answered by
Damon
In the end for the carbon
3 C --> CO + C + CaCO3
so in the end we need 3 C atoms for every molecule of CaCO3
C = 12 g
Ca = 40 g
3O = 3*16 = 48 g
so CaCO3 = 100 g/mol
so we get 7.90 mols of CaCO3
for that we need 3*7.9 = 23.7 mols of C
23.7 * 12 = 284.4 grams of Carbon
3 C --> CO + C + CaCO3
so in the end we need 3 C atoms for every molecule of CaCO3
C = 12 g
Ca = 40 g
3O = 3*16 = 48 g
so CaCO3 = 100 g/mol
so we get 7.90 mols of CaCO3
for that we need 3*7.9 = 23.7 mols of C
23.7 * 12 = 284.4 grams of Carbon
Answered by
Damon
we got carbon as a result of step 2 so we could put it back in to step 1
then we would only need two carbons fo each CaCO3
then we would only need two carbons fo each CaCO3
Answered by
Damon
2 * 7.9 = 15.8 mols C
15.8 * 12 g/mol = 189.6 grams of carbon
15.8 * 12 g/mol = 189.6 grams of carbon