Question

When heated, calcium carbonate decomposes to yield calcium oxide and carbon dioxide gas via the reaction
CaCO3 ---> CaO + CO2

How many grams of calcium carbonate are needed to produce 73.0 of carbon dioxide at STP?

Answer 1

I did it like

76.0Lof CO2/22.4L = 3.2589moles of CO2
Moles of CO2 = Moles of CaCaCO3

to convert it in g's:
3.2589moles of CaCO3*(100g/1mole Of CaCO3)
= 325.89g

Answer 2

73 g CO2 is how many moles?

From the balanced equation, it takes one mole of calcium carbonate to produced one mole of carbon dioxide.

Answer 3

ohh yea i figured tht out. nd thnxxx a lot fohh ur help.

Answer 4

73.0 WHAT</b) of carbon dioxide?????
I will assume you mean grams but if not grams you must adjust the work accordingly.
You have the balanced equation.
2. Convert 73.0 g CO2 to mols. moles = grams/molar mass.
3. Using the coefficients in the balanced equation, convert moles CO2 to moles CaCO3.
4. Now convert moles CaCO3 to grams. grams CaCO3 = moles CaCO3 x molar mass CaCO3.

Answer 5

OK. So it's 73.0 LITERS. You made a typo in your work where you typed 76.0 BUT your answer is correct for 73.0/22.4. If you are keying this answer into an on-line database, I suspect your answer is coming back wrong because you are posting too many significant figures. Your answer should be rounded to 326 grams CaCO3.