The anesthetic chloroform (CHCI3, molar mass=119.36 g/mol) can be prepared by the reaction of methane CH4 molar mass= 16.04 g/mol with chlorine according to the following balanced equation: CH4+3CI2 CHCI3+3HCI
Assuming 10.50g of methane and 20.50g of chlorine are used, answer the following questions:
How many moles of chloroform are formed? Be sure your answer has the correct number of significant figures.
moles of chloroform
To determine the moles of chloroform formed, we first need to identify the limiting reactant.
1. Calculate the moles of methane used:
moles of CH4 = mass / molar mass
moles of CH4 = 10.50g / 16.04 g/mol
moles of CH4 ≈ 0.654 mol
2. Calculate the moles of chlorine used:
moles of Cl2 = mass / molar mass
moles of Cl2 = 20.50g / (35.45 g/mol + 35.45 g/mol)
moles of Cl2 ≈ 0.289 mol
3. Determine the limiting reactant:
Since the balanced equation shows that 1 mole of methane reacts with 3 moles of chlorine, we need to determine how many moles of chlorine would react with the moles of methane determined.
moles of Cl2 needed = (moles of CH4) x 3
moles of Cl2 needed ≈ 0.654 mol x 3 ≈ 1.963 mol
Since we only have 0.289 mol of Cl2, it is the limiting reactant.
4. Calculate the moles of chloroform formed using the limiting reactant:
moles of CHCl3 = moles of limiting reactant
moles of CHCl3 ≈ 0.289 mol
Therefore, approximately 0.289 moles of chloroform are formed.