How many milliliters of 0.200M HCl can react with 3.25g CaCO3?
To determine the number of milliliters of 0.200M HCl required to react with 3.25g CaCO3, we need to follow these steps:
Step 1: Write the balanced chemical equation for the reaction.
CaCO3 + 2HCl → CaCl2 + CO2 + H2O
Step 2: Calculate the molar mass of CaCO3.
Ca: 1 * 40.08 g/mol = 40.08 g/mol
C: 1 * 12.01 g/mol = 12.01 g/mol
O: 3 * 16.00 g/mol = 48.00 g/mol
Total molar mass of CaCO3 = 40.08 + 12.01 + 48.00 = 100.09 g/mol
Step 3: Determine the number of moles of CaCO3.
Number of moles = mass / molar mass
Number of moles of CaCO3 = 3.25 g / 100.09 g/mol = 0.0325 mol
Step 4: Determine the number of moles of HCl required for the reaction.
From the balanced equation, we can see that the stoichiometric ratio between CaCO3 and HCl is 1:2. This means that for every mole of CaCO3, 2 moles of HCl are needed.
Number of moles of HCl = 2 * 0.0325 mol = 0.065 mol
Step 5: Calculate the volume of 0.200M HCl required.
Molarity (M) = moles / volume (in liters)
Rearranging the formula: volume (in liters) = moles / Molarity
Volume (in liters) = 0.065 mol / 0.200 mol/L = 0.325 L
Step 6: Convert the volume from liters to milliliters.
1 L = 1000 mL
Volume (in mL) = 0.325 L * 1000 mL/L = 325 mL
Therefore, 325 milliliters of 0.200M HCl is required to react with 3.25g of CaCO3.
balance the equation:
2HCl + CaCO3 >> CO2 + H2O + CaCl2
figure the moles of calcium carbonate:
3.25/molmassCaCO3
then you must need twice that of acid.
2*3.25/molmassCaCO3 = volume*.2M
solvefor volume in liters. change to milliliters.