Question

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?

Answer 1

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