How much pure calcium carbonate (100.09 g/mol) is required to produce 40.0 mL carbon dioxide gas according to the reaction used in this experiment?
Your unknown will contain between 25-100% calcium carbonate. Answer to 3 significant figures. Do not include the unit, g.
Hint: ( 0.0400 L CO2 ) X ( 1 mol CO2 / 22.4 L ) X ( __ mol CaCO3 / __ mol CO2 ) X ( __ g CaCO3 / 1 mol CaCO3 )
0.178
To find out how much pure calcium carbonate is required to produce 40.0 mL of carbon dioxide gas, we can use stoichiometry and the given molar mass for calcium carbonate.
First, let's set up the information we have:
1. The volume of carbon dioxide gas produced is 40.0 mL.
2. The molar mass of carbon dioxide is 44.01 g/mol (not provided in the question, but commonly known).
3. The molar mass of calcium carbonate is 100.09 g/mol (given in the question).
Next, we can use stoichiometry to convert the volume of carbon dioxide gas into the moles of carbon dioxide gas produced. The conversion factor is based on the ideal gas law at standard temperature and pressure (STP), where 1 mole of any gas occupies 22.4 liters.
So, we start with the given volume of carbon dioxide gas:
40.0 mL CO2
Now, we can convert this volume into moles of carbon dioxide using the conversion factor:
(40.0 mL CO2) x (1 mol CO2 / 22.4 L) = 1.7857 mol CO2 (rounded to 4 decimal places)
Now that we have the moles of carbon dioxide, we can use stoichiometry to find the moles of calcium carbonate. The balanced chemical equation for the reaction must be known for this step, but it is not provided in the question. Without the balanced equation, it's impossible to determine the conversion factor.
Once we have the moles of calcium carbonate, we can use its molar mass to find the mass of calcium carbonate needed. As given in the question, the molar mass of calcium carbonate is 100.09 g/mol.
Finally, multiply the moles of calcium carbonate by its molar mass to find the mass:
(moles of CaCO3) x (molar mass of CaCO3) = mass of CaCO3
Unfortunately, without the balanced chemical equation, we cannot proceed to calculate the mass of calcium carbonate required.