What mass of carbon dioxide will be produced?
(Calcium Chloride and water are also produced)
A second question using the previous question:
If only 3.0g of carbon dioxide is produced,instead of the previously calculated amount, what is the percentage yield?
Answer:
CaCO3+2HCl--> CaCl2+H2O+CO2
10 g CaCO3/100 g per mole CaCO3 gives .1 mole CaCO3
.1 mole CaCO3 (1 mole CO2/ 1 mole CaCO3) = .1 mole CO2
.1mole CO2*44 g per mole CO2 gives 4.4 g CO2
percent yield is
actual yield divided by theoretical yield all multiplied by 100%
3g/4.4gx100%=68.12%
CaCO3 + 2 HCl = CO2 + CaCl2 + H2O (balanced)
You've got 10.0 g of CaCO3
MW = 100, so that's 0.1 mol
You've got 8 g of HCl
MW = 36, so that's 0.22 mol
0.1 mol of CaCO3 would react with 0.2 mol of HCl, so HCl is in excess and CaCO3 is limiting.
0.1 mol of CaCO3 will produce 0.1 mol of CO2.
MW = 44, so that's 4.4 g of CO2 produced.
You actually formed 3.0 g, so your yield is
3.0 / 4.4 * 100% = 68%
You need to read your book - it is a simple matter of conversion between moles and mass. This is elementary - as time goes on you are going to get harder questions. Who is gonna pass the course? You or us?
Ron.
1. write out the equation first (CaCO3 + HCl ---> CO2 + CaCl + H2O); balance it
2. find the limiting reactant (calcium carbonate or HCl)
3. use the limiting reactant (in moles) to find mass of carbon dioxide produced (in grams) because your question asks for the mass
4. take your answer from #3
( grams of CO2 produced / 3.0g ) x 100 = percent yield
goodluck!
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