The question posted under Joe says calcium bicarbonate and that's what I answered. For CaCO3 is is a different equation but the process is exactly the same.
Step 1. Write and balance the equation.
CaCO3 + H2SO4 ==> CaSO4 + H2O + CO2
Step 2. How many mols do you need to neutralize; i.e., how mols is in 10 mL of 3M H2SO4.
mols = M x L. You know M and you know L. Solve for mols H2SO4.
Step 3. Using the coefficients in the balanced equation, convert mols H2SO4 to mols CaCO3.
?mols H2SO4 x (1 mol CaCO3/1 mol H2SO4) = ?mols H2SO4 x 1/1 so mols H2SO4 = mols CaCO3.
Step 4. Now convert mols CaCO3 to grams CaCO3.
g CaCO3 = mols CaCO3 x molar mass CaCO3.
This four-step procedure will work almost any stoichiometry problem in the book. I didn't write anything different here than I did with your Ca(HCO3)2 post, except of course I changed to CaCO3. If you don't understand a step or the reasoning behind the step explain what you don't understand and I can help you through that. This is as simple as it gets.
i want to neutralize a 10 ml 3 molar sulfuric acid using calcium carbonate but don't know how to do the math the explanation confused me.
(also i have to use sig fig and uncertainty)
1 answer
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