Consider the following reaction:
CaCN2 + 3H2O �¨ CaCO3 + 2NH3
105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency, which reactant is in excess and how much is leftover? The molar mass of CaCN2 is 80.11 g/mol. The
molar mass of CaCO3 is 100.09 g/mol.
1. H2O; 70.8 g left over
2. H2O; 10.7 g left over
3. CaCN2; 10.7 g left over
4. H2O; 7.20 g left over
5. CaCN2; 7.20 g left over
6. CaCN2; 70.8 g left over
I'm not getting any of these answers. I think H20 is in excess, but I get about 54 g for the amount.
You're right. Water is in excess and a quick calculation I get 7.15 remaining; therefore, I would pick 4 a the answer.
Is that CaCN2 or Ca(CN)2. That may be your problem.
CaCN2
H2O; 7.20 g left over
Well, it seems like you're in a bit of a H2O-muddle here. Let me help you clear the confusion with a touch of humor!
To solve this question, we'll want to compare the number of moles of CaCN2 and H2O available for reaction.
Moles of CaCN2 = 105.0 g / 80.11 g/mol = 1.310 mol
Moles of H2O = 78.0 g / 18.02 g/mol = 4.33 mol
Now, let's look at the coefficients in front of CaCN2 and H2O in the balanced equation - 1 and 3, respectively.
Since 1 mole of CaCN2 reacts with 3 moles of H2O, for the given amounts, 1.310 moles of CaCN2 would need 3 * 1.310 = 3.93 moles of H2O.
We have 4.33 moles of H2O available, which is more than enough to react with 1.310 moles of CaCN2. Therefore, H2O is in excess.
To find the amount of excess H2O, we subtract the amount used in the reaction from the total amount available:
Excess H2O = Moles of H2O available - Moles of H2O used in reaction
Excess H2O = 4.33 mol - (3 * 1.310 mol) = 0.400 mol
Now, let's convert this excess amount into grams:
Excess H2O (g) = Moles of H2O (excess) * Molar mass of H2O
Excess H2O (g) = 0.400 mol * 18.02 g/mol = 7.208 g
So, it looks like option 4 is the correct answer. The excess reactant is H2O, and approximately 7.20 g of H2O will be left over.
Keep smiling and crunching those chemistry numbers!
To determine which reactant is in excess and how much is leftover, we need to compare the number of moles of each reactant and their stoichiometric ratio in the balanced chemical equation.
First, let's calculate the number of moles for each reactant:
Moles of CaCN2 = mass of CaCN2 / molar mass of CaCN2
Moles of CaCN2 = 105.0 g / 80.11 g/mol
Moles of CaCN2 ≈ 1.31 mol
Moles of H2O = mass of H2O / molar mass of H2O
Moles of H2O = 78.0 g / 18.015 g/mol
Moles of H2O ≈ 4.33 mol
Next, we need to determine the stoichiometric ratio of the reactants. According to the balanced equation:
CaCN2 + 3H2O → CaCO3 + 2NH3
The ratio of CaCN2 to H2O is 1:3. This means that for every 1 mole of CaCN2, we need 3 moles of H2O.
Now, let's compare the moles of each reactant to the stoichiometric ratio:
Moles of CaCN2 / Stoichiometric ratio = 1.31 mol / 1 = 1.31 mol
Moles of H2O / Stoichiometric ratio = 4.33 mol / 3 = 1.44 mol
Since the stoichiometric ratio requires more moles of H2O (3:1), it indicates that CaCN2 is the limiting reactant, and H2O is in excess.
To determine the amount of excess reactant leftover, we need to subtract the moles of the limiting reactant.
Excess moles of H2O = Moles of H2O - (Stoichiometric ratio * Moles of CaCN2)
Excess moles of H2O = 4.33 mol - (3 * 1.31 mol)
Excess moles of H2O ≈ 0.4 mol
Now, let's calculate the mass of the excess H2O:
Mass of excess H2O = Excess moles of H2O * Molar mass of H2O
Mass of excess H2O = 0.4 mol * 18.015 g/mol
Mass of excess H2O ≈ 7.20 g
Therefore, the correct answer is option 4. H2O is in excess, and there is approximately 7.20 g of H2O left over.