How many grams of calcium chloride will be produced when 30.0g of calcium carbonate are combined with 14.0g of hydrochloric acid?
To determine the number of grams of calcium chloride produced when calcium carbonate and hydrochloric acid react, we first need to write and balance the chemical equation:
CaCO3 + 2 HCl → CaCl2 + H2O + CO2
From the balanced equation, we can see that 1 mole of calcium carbonate (CaCO3) reacts with 2 moles of hydrochloric acid (HCl) to produce 1 mole of calcium chloride (CaCl2).
1 mole of CaCO3 has a molar mass of 100.09 g/mol, so 30.0g of CaCO3 is equal to:
30.0 g / (100.09 g/mol) = 0.2996 moles of CaCO3
2 moles of HCl have a molar mass of 2 * 36.46 g/mol = 72.92 g/mol, so 14.0g of HCl is equal to:
14.0 g / (72.92 g/mol) = 0.1919 moles of HCl
Since the reaction ratio is 1:1 between CaCO3 and CaCl2, the number of moles of CaCl2 produced is also equal to 0.1919 moles.
The molar mass of CaCl2 is 110.98 g/mol, so the mass of calcium chloride produced is:
0.1919 moles * (110.98 g/mol) = 21.3 grams of calcium chloride
Therefore, 21.3 grams of calcium chloride will be produced when 30.0 grams of calcium carbonate are combined with 14.0 grams of hydrochloric acid.
To determine the number of grams of calcium chloride produced when 30.0g of calcium carbonate reacts with 14.0g of hydrochloric acid, we need to calculate the balanced chemical equation for the reaction and use stoichiometry.
The balanced chemical equation is:
CaCO3 + 2HCl -> CaCl2 + CO2 + H2O
From the equation, we can see that one mole of calcium carbonate (CaCO3) reacts with two moles of hydrochloric acid (HCl) to produce one mole of calcium chloride (CaCl2).
Step 1: Convert the given masses of calcium carbonate and hydrochloric acid to moles.
Molar mass of calcium carbonate (CaCO3) = 40.08 g/mol (40.08 g/mol/mol)
Molar mass of hydrochloric acid (HCl) = 36.46 g/mol (36.46 g/mol/mol)
Moles of calcium carbonate (CaCO3) = given mass / molar mass
Moles of calcium carbonate = 30.0 g / 40.08 g/mol = 0.748 mol (rounded to three decimal places)
Moles of hydrochloric acid (HCl) = given mass / molar mass
Moles of hydrochloric acid = 14.0 g / 36.46 g/mol = 0.384 mol (rounded to three decimal places)
Step 2: Determine the limiting reactant.
To find the limiting reactant, compare the mole ratio of calcium carbonate to hydrochloric acid with the actual ratio obtained from the calculations. The reactant that produces the lesser amount of product is the limiting reactant.
From the balanced equation, the mole ratio of CaCO3 to HCl is 1:2. This means that for every mole of calcium carbonate, we need two moles of hydrochloric acid.
Adjusted mole ratio:
CaCO3 : HCl = 1 : 2
Actual mole ratio:
CaCO3 : HCl = 0.748 mol : 0.384 mol (rounded to three decimal places)
Since the actual mole ratio of calcium carbonate to hydrochloric acid is less than the adjusted mole ratio, hydrochloric acid is the limiting reactant.
Step 3: Calculate the moles of calcium chloride produced.
From the balanced equation, we can see that one mole of calcium carbonate produces one mole of calcium chloride. Therefore, the moles of calcium chloride produced will be the same as the moles of hydrochloric acid used since they are in a 1:1 ratio.
Moles of calcium chloride = moles of hydrochloric acid = 0.384 mol (from earlier calculations)
Step 4: Convert moles of calcium chloride to grams.
Molar mass of calcium chloride (CaCl2) = 110.98 g/mol (110.98 g/mol/mol)
Mass of calcium chloride = moles of calcium chloride × molar mass
Mass of calcium chloride = 0.384 mol × 110.98 g/mol = 42.637 g (rounded to three decimal places)
Therefore, approximately 42.637 grams of calcium chloride will be produced when 30.0 grams of calcium carbonate are combined with 14.0 grams of hydrochloric acid.
This is a limiting reagent (LR) problem and you know that because amounts are given for BOTH reactants.
CaCO2 + 2HCl ==> H2O + CO2 + CaCl2
mols CaCO3 = grams/molar mass
mols HCl = grams/molar mass
Using the coefficients in the balanced equation, convert mols CaCO3 to mols CaCl2.
Do the same to convert mols HCl to g CaCl2.
It is likely that these two values will not agree which means one of them is wrong; the correct value in LR problems is ALWAYS the smaller value and the reagent responsible for that number is the LR.
Now convert mols CaCl2 to grams. g = mols x molar mass