How do I work the following question? I found the amu of CaCO3=101.0872g/mol; 2HCL=72.92128g/mol; H2O=18.1582g/mol;CO2=44.0098g/mol; and, CaCl2=110.9834g/mol.
You carefully weigh out 10.00g of powder and add it to 40.50g of solution. You notice bubbles as a reaction takes place. You then weigh the resulting solution and find that it has a mass of 46.40g . The relevant equation is
CaCO_3+2HCL(aq) -> H_2O(l)+CO_2+CaCl_2
What mass of CO_2 was produced in this reaction?
You started with 50.50 of reactants, so you get that same mass of products. Subtract the mass of the solution, and you have the mass of the gas.
To solve this question, you need to use the concept of stoichiometry and the given information.
First, let's find the moles of CaCO3 and HCl using their given molar masses:
Molar mass of CaCO3 = 101.0872 g/mol
Molar mass of HCl = 72.92128 g/mol
Moles of CaCO3 = mass / molar mass
= 10.00 g / 101.0872 g/mol
Moles of HCl = mass / molar mass
= 40.50 g / 72.92128 g/mol
Next, to determine the limiting reactant, we compare the moles of CaCO3 and HCl using the stoichiometric coefficients in the balanced equation.
From the balanced equation: CaCO3 + 2HCl(aq) -> H2O(l) + CO2 + CaCl2,
we see that 1 mole of CaCO3 reacts with 2 moles of HCl to produce 1 mole of CO2.
So, if we have X moles of CaCO3, we need 2X moles of HCl to react completely.
Now, compare the moles of CaCO3 and HCl:
If Moles of CaCO3 < Moles of HCl / 2, then CaCO3 is the limiting reactant.
If Moles of CaCO3 > Moles of HCl / 2, then HCl is the limiting reactant.
Once you identify the limiting reactant, you can determine the moles of CO2 produced by the balanced equation.
Finally, calculate the mass of CO2 produced using the molar mass of CO2:
Mass of CO2 = moles of CO2 * molar mass of CO2
By following these steps, you should be able to determine the mass of CO2 produced in this reaction.