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
First, we need to determine the limiting reactant by calculating the moles of methane and chlorine used in the reaction:
Moles of methane = 10.50g / 16.04 g/mol = 0.654 moles
Moles of chlorine = 20.50g / 70.90 g/mol = 0.289 moles
Since there are 3 moles of chlorine required for every mole of methane in the reaction, we can see that chlorine is the limiting reactant as it is present in a smaller amount.
Now, we can calculate the moles of chloroform formed using the mole ratio from the balanced equation:
1 mole of methane produces 1 mole of chloroform
0.289 moles of chlorine * (1 mole of chloroform / 3 moles of chlorine) = 0.0963 moles of chloroform
Therefore, 0.0963 moles of chloroform are formed.