According to the following equation:
CaCO3 + 2 HCl---> CaCl2 + H20 + CO2
How many grams of CO2 can be produced from the reaction of 20.0 mL of 0.400M HCl solution with excess CaCO3?
To determine the number of grams of CO2 produced, we need to use stoichiometry and dimensional analysis.
Step 1: Calculate the number of moles of HCl used.
Given that the volume of the HCl solution is 20.0 mL and the concentration is 0.400 M, we can use the formula:
Molarity (M) = moles/volume (L)
0.400 M = moles/0.0200 L
Solving for moles, we get:
moles of HCl = 0.400 M x 0.0200 L = 0.00800 moles
Step 2: Use the balanced equation to determine the stoichiometric ratio between HCl and CO2.
From the balanced equation, we see that 1 mole of CaCO3 reacts with 2 moles of HCl to produce 1 mole of CO2. Therefore, there is a 1:1 ratio of moles between HCl and CO2.
Step 3: Calculate the moles of CO2 produced.
Since the number of moles of HCl is equal to the number of moles of CO2, we have:
moles of CO2 = moles of HCl = 0.00800 moles
Step 4: Convert moles of CO2 to grams.
To convert moles to grams, we need to use the molar mass of CO2, which is 44.01 g/mol.
grams of CO2 = moles of CO2 x molar mass of CO2
grams of CO2 = 0.00800 moles x 44.01 g/mol = 0.3528 grams
Therefore, approximately 0.3528 grams of CO2 can be produced from the reaction.