16 gram of CaC2 and 18 gram of H2O were mixed in the given reaction,
CaC2 + 2H2O it gives C2H2 + Ca(OH)2.
A, what mass of Ca(OH)2 is formed?
B, How many grams of the excess reactant left unreacted?
To find out how many moles of CaC2 and H2O are present, we can use the formula:
moles = mass/molecular weight
The molecular weight of CaC2 = 40.08 (Ca) + 2 x 12.01 (C) = 64.1 g/mol
The molecular weight of H2O = 2 x 1.01 (H) + 16 (O) = 18.02 g/mol
Now, we calculate the moles of CaC2 and H2O:
moles of CaC2 = 16 g / 64.1 g/mol = 0.25 mol
moles of H2O = 18 g / 18.02 g/mol = 1 mol
From the given reaction, we know that 1 mole of CaC2 reacts with 2 moles of H2O to form 1 mole of Ca(OH)2.
Now, let's check for the limiting reactant:
0.25 mol CaC2 (1 mol H2O / 1 mol CaC2) = 0.25 mol H2O is required.
Since there are 1 mol of H2O present (which is more than what is required), CaC2 is the limiting reactant.
Now, we can calculate the mass of Ca(OH)2 formed:
0.25 mol CaC2 (1 mol Ca(OH)2 / 1 mol CaC2) = 0.25 mol Ca(OH)2 is formed
The molecular weight of Ca(OH)2 = 40.08 (Ca) + 2(16 + 1.01) = 74.1 g/mol
Mass of Ca(OH)2 formed = 0.25 mol x 74.1 g/mol = A. 18.53 grams
Now, let's calculate the amount of excess reactant remaining:
1 mol H2O (initial) - 0.25 mol H2O (reacted) = 0.75 mol H2O (remaining)
Mass of H2O remaining = 0.75 mol x 18.02 g/mol = B. 13.52 grams
To answer these questions, we'll need to calculate the molar masses of the compounds involved and determine the limiting reactant.
A. Let's calculate the mass of Ca(OH)2 formed:
1. Calculate the number of moles of CaC2:
Molar mass of CaC2 = atomic mass of Ca + 2 * atomic mass of C = 40.08 + 2 * 12.01 = 64.10 g/mol
Number of moles of CaC2 = mass of CaC2 / molar mass of CaC2 = 16 g / 64.10 g/mol = 0.25 mol
2. Calculate the number of moles of H2O:
Molar mass of H2O = 2 * atomic mass of H + atomic mass of O = 2 * 1.01 + 16.00 = 18.02 g/mol
Number of moles of H2O = mass of H2O / molar mass of H2O = 18 g / 18.02 g/mol = 0.999 mol (approximately 1 mol)
3. The balanced equation tells us that 1 mol of CaC2 reacts with 2 mol of H2O to produce 1 mol of Ca(OH)2. Therefore, the mole ratio of CaC2 to Ca(OH)2 is 1:1.
Since we have 0.25 mol of CaC2, we expect to produce 0.25 mol of Ca(OH)2.
4. Calculate the mass of Ca(OH)2:
Molar mass of Ca(OH)2 = atomic mass of Ca + atomic mass of O + 2 * atomic mass of H = 40.08 + 16.00 + 2 * 1.01 = 74.10 g/mol
Mass of Ca(OH)2 = number of moles of Ca(OH)2 * molar mass of Ca(OH)2 = 0.25 mol * 74.10 g/mol = 18.52 g
Therefore, the mass of Ca(OH)2 formed is 18.52 grams.
B. To determine the mass of the excess reactant left unreacted, we first need to identify the limiting reactant.
The mole ratio between CaC2 and H2O is 1:2, meaning that for every mole of CaC2, we need 2 moles of H2O to react. Since we have 0.25 mol of CaC2, it would require 2 * 0.25 = 0.5 mol of H2O to react completely.
However, we only have 1 mol of H2O available, which is more than enough to react with the CaC2. Therefore, H2O is in excess, and CaC2 is the limiting reactant.
To calculate the mass of the excess reactant left unreacted (H2O), we can subtract the amount of H2O consumed in the reaction from the initial amount.
The molar mass of H2O is 18.02 g/mol. We used 0.25 mol of CaC2, which means 0.25 mol * 2 mol H2O/1 mol CaC2 = 0.5 mol H2O reacted.
Initial amount of H2O = 18 g
Amount of H2O consumed = 0.5 mol * 18.02 g/mol = 9.01 g
Mass of excess H2O = Initial amount - Amount consumed = 18 g - 9.01 g = 8.99 g (approximately)
Therefore, approximately 8.99 grams of the excess reactant (H2O) is left unreacted.