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Sagot :
If HCl was completely used up during the reaction, CO2 is probably the excess reactant. However, let's perform the calculations to be sure.
- First, let's transform grams into moles of each compound. For this, we need the molar mass of CO2 and HCl.
Molar mass:
CO2 = (12x1) + (16x2) = 44 g/mol
HCl = (1x1) + (35.45x1) = 36.45 g/mol
For CO2:
44 g ---- 1 mol
15.3 g ---- x mol
x = 0.3478 mol of CO2
For HCl:
36.45 g --- 1 mol
4.03 g --- x mol
x = 0.1110 mol of HCl
- Now let's use the equation ratio to find out which one is the excess reactant and which one is the limiting reagent.
CO2 + 4 HCl --> CCl4 + 2 H2O
The reaction gives us that 1 mol of CO2 reacts with 4 mol of HCl.
So if we have 0.3478 mol of CO2:
1 mol CO2 ---- 4 mol HCl
0.3478 mol of CO2 --- x
x = 1.39
If we have 0.1110 mol of HCl
1 mol CO2 ---- 4 mol HCl
x mol of CO2 ---- 0.1110 mol of HCl
x = 0.027 mol of CO2
As we can see, we should have more HCl than we actually have to react with the quantity of CO2. So the reagent in excess is CO2.
Answer: Alternative "B" CO2
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