a sample of limestone and other solid materials is heated and the limestone decomposes to give calcium oxide and carbon dioxide:

CaCO3 ----> CaO + CO2

a 1.506 g sample of limestone containing material produces 284 mL of CO2 at STP. what is the mass percent os CaCO3 in the original sample?
PLEASE HELP

change the volume of carbon dioxide to moles of CO2, and you have the same number of moles of CaO. Thence, you had the same number of moles of limestone, change that to massCaCO3

percent= massyoufigured above/1.506

so then would i take:

.284 L/44 g/mol =.006
then .006*100 g/mol =.645
then take .645/1.506 = .429
---->.429 * 100 = 42.9%
is this how i would get the answer?

To find the mass percent of CaCO3 in the original sample, we need to determine the mass of CaCO3 in the sample and then calculate its percentage in relation to the total mass of the sample.

First, let's calculate the molar mass of CaCO3:
- The atomic mass of calcium (Ca) is 40.08 g/mol.
- The atomic mass of carbon (C) is 12.01 g/mol.
- The atomic mass of oxygen (O) is 16.00 g/mol.
So, the molar mass of CaCO3 is: (1 × 40.08) + (1 × 12.01) + (3 × 16.00) = 100.09 g/mol.

Next, we can calculate the number of moles of CO2 produced. We can use the ideal gas law to find the number of moles:
PV = nRT, where P is the pressure (STP = 1 atm), V is the volume (284 mL = 0.284 L), n is the number of moles, R is the ideal gas constant (0.0821 L·atm/(mol·K)), and T is the temperature (273 K for STP).
Therefore, n = (P × V) / (R × T) = (1 atm × 0.284 L) / (0.0821 L·atm/(mol·K) × 273 K) ≈ 0.011 moles.

Since the balanced equation shows that 1 mole of CaCO3 produces 1 mole of CO2, we can conclude that 0.011 moles of CaCO3 decomposed.

Finally, we can calculate the mass of CaCO3 in the sample:
mass of CaCO3 = moles of CaCO3 × molar mass of CaCO3 = 0.011 moles × 100.09 g/mol ≈ 1.10 g.

Now, we can find the mass percent of CaCO3 in the original sample:
mass percent of CaCO3 = (mass of CaCO3 / mass of original sample) × 100%
= (1.10 g / 1.506 g) × 100% ≈ 73.0%.

Therefore, the mass percent of CaCO3 in the original sample is approximately 73.0%.