Calcium carbonate was reacted with hydrogen chloride. What is the percentage yield of carbon(iv)oxide produced if 3.65g of the gas is collected when 10g of calcium(iii)oxocarbonate reacts
First there is no such thing as calcium(iii) oxycarbonate. Ditto for calcium(iv) oxide.
CaCO3 + 2HCl ==> H2O + CO2 + CaCl2
mols CaCO3 = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols CaCO3 to mols CaO.
Convert mols CaO to grams CaO by grams = mols x molar mass = ?. This is the theoretical yield (TY). The actual yield (AY) given in the problem is 3.65.
%yield = (AY/TY)*100 = ?
To calculate the percentage yield, you need to compare the actual yield (the amount of carbon(iv)oxide produced) to the theoretical yield (the amount of carbon(iv)oxide that would be produced when the reaction goes to completion).
First, let's write the balanced equation for the reaction:
CaCO3 + 2 HCl -> CaCl2 + CO2 + H2O
From the balanced equation, we can see that 1 mole of calcium carbonate (CaCO3) reacts with 2 moles of hydrogen chloride (HCl) to produce 1 mole of carbon(iv)oxide (CO2).
Now, we need to calculate the theoretical yield of carbon(iv)oxide. To do this, we'll use stoichiometry and the molar mass of calcium carbonate (100.09 g/mol) to find the number of moles of calcium carbonate used in the reaction:
moles of CaCO3 = mass of CaCO3 / molar mass of CaCO3
moles of CaCO3 = 10 g / 100.09 g/mol
moles of CaCO3 = 0.1 mol
According to the balanced equation, we know that 1 mole of calcium carbonate produces 1 mole of carbon(iv)oxide. Therefore, the theoretical yield of carbon(iv)oxide would be equal to the number of moles of calcium carbonate used in the reaction.
theoretical yield of CO2 = moles of CaCO3 = 0.1 mol
Next, we need to calculate the actual yield of carbon(iv)oxide. We are given that 3.65g of the gas is collected, so the actual yield of carbon(iv)oxide is 3.65 g.
Now we can calculate the percentage yield using the formula:
percentage yield = (actual yield / theoretical yield) x 100
percentage yield = (3.65 g / 0.1 mol) x 100
percentage yield = 36.5%
Therefore, the percentage yield of carbon(iv)oxide produced in this reaction is 36.5%.