lime, Ca(OH)2 can be used to neutralize an acid spill.A 5.06-g sample of Ca(OH)reacts with excess of hydrochloric acid:6.74 g of calcium chloride is collected what is the percent yield of this experiment

Ca(OH)2 + 2HCI ====> CaCl2 + 2H2O

mols Ca(OH)2 = grams/molar mass = ?

mols CaCl2 = mols Ca(OH)2 from the equation (1 mol Ca(OH)2 = 1 mol CaCl2)
g CaCl2 = mols CaCl2 x molar mass CaCl2.
This is the theoretical yield (TY). The actual yield (AY) is given in the problem as 6.74 g.
%yield = (AY/TY)*100 = ?

To calculate the percent yield, we need to compare the actual yield to the theoretical yield.

First, let's calculate the theoretical yield of calcium chloride (CaCl2) using the given information:

Molar mass of Ca(OH)2 = 74.09 g/mol
Molar mass of CaCl2 = 110.98 g/mol

To find the theoretical yield, we need to convert the given mass of Ca(OH)2 to moles.
Moles of Ca(OH)2 = mass / molar mass
Moles of Ca(OH)2 = 5.06 g / 74.09 g/mol
Moles of Ca(OH)2 ≈ 0.0683 mol

According to the balanced equation, the ratio of moles of Ca(OH)2 to moles of CaCl2 is 1:1. Therefore, the moles of CaCl2 produced should also be 0.0683 mol.

To find the mass of CaCl2 produced, we need to use the moles of CaCl2 and the molar mass of CaCl2:
Mass of CaCl2 = moles × molar mass
Mass of CaCl2 = 0.0683 mol × 110.98 g/mol
Mass of CaCl2 ≈ 7.57 g

Now we can calculate the percent yield:

Percent yield = (Actual yield / Theoretical yield) × 100

Actual yield = 6.74 g (given)

Percent yield = (6.74 g / 7.57 g) × 100
Percent yield ≈ 88.9%

Therefore, the percent yield of this experiment is approximately 88.9%.

To calculate the percent yield of a chemical reaction, you need to compare the actual amount of product obtained with the theoretical amount of product that could be obtained.

First, let's calculate the theoretical yield of calcium chloride (CaCl2) based on the balanced equation:
1 mole of Ca(OH)2 reacts with 2 moles of HCI to produce 1 mole of CaCl2.

1 mole of Ca(OH)2 is equal to its molar mass:
Ca(OH)2 = 40.08 g/mol (atomic mass of Ca) + 2 * (16.00 g/mol + 1.01 g/mol) (atomic mass of O and H) = 74.08 g/mol

Based on the given information, we have a 5.06 g sample of Ca(OH)2, so we can calculate the number of moles of Ca(OH)2:
moles of Ca(OH)2 = mass of Ca(OH)2 / molar mass of Ca(OH)2
moles of Ca(OH)2 = 5.06 g / 74.08 g/mol ≈ 0.0684 mol

Since the reaction between Ca(OH)2 and HCI is in a 1:2 ratio, the moles of CaCl2 produced will be twice the moles of Ca(OH)2:
moles of CaCl2 = 2 * moles of Ca(OH)2
moles of CaCl2 = 2 * 0.0684 mol ≈ 0.1368 mol

Now let's calculate the theoretical mass of CaCl2:
mass of CaCl2 = moles of CaCl2 * molar mass of CaCl2
mass of CaCl2 = 0.1368 mol * (40.08 g/mol + 2 * 35.45 g/mol) ≈ 13.11 g

So, the theoretical yield of calcium chloride is approximately 13.11 g.

The percent yield is calculated by dividing the actual yield by the theoretical yield and then multiplying by 100:
percent yield = (actual yield / theoretical yield) * 100

From the given information, the actual yield of calcium chloride is 6.74 g.

percent yield = (6.74 g / 13.11 g) * 100
percent yield ≈ 51.39%

Therefore, the percent yield of this experiment is approximately 51.39%.