A 1.056 sample of limestone produces 284 mL of CO2 @STP. what is the mass % of CaCO3 is in the original sample?
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 of CaCO3 is in the original sample?
Please show work, I have been having a really hard time with this. Thanks
- Trevor HLv 79 years agoFavorite Answer
So far , quite a number of different answers - can it be that difficult?
CaCO3 → CaO + CO2
1mol CaCO3 produces 1 mol CO2
At STP, 1 mol CO2 = 22.4 litres
At STP , 0.284 L CO2 = 0.284/22.4 = 0.01268 mol CO2
This came from decomposition of 0.01268 mol CaCO3
Molar mass CaCO3 = 100g/mol
0.01268 mol = 0.01268*100 = 1.268g CaCO3 in sample
% CaCO3 in original sample = 1.268/1.506 *100 = 84.2% CaCO3 in sample.
Now at least you have 2 correct answers that agree and ( at last count) 3 incorrect answers.
- 9 years ago
First of all we need to look at the decomposition equation of limestone. We see that calcium carbonate (calcite) decomposes in a 1:1 ratio to CaO and CO2 - i.e. for every mole of CaCO3 you decompose you will produce 1 mole of CaO and 1 mole of CO2.
Knowing this we can use the number of moles of CO2 produced as evidence for the number of moles CaCO3 in the original sample (because these values will be the same). If we can assume that CO2 is an ideal gas (usually the case), we can say:
PV = nRT (this is the ideal gas equation)
P = pressure, V = volume, n = Number of Moles (what we will solve for), R = ideal gas constant, T = temperature. At STP we know that P = 100,000 Pa and T = 273.15 K, and we are given volume and R is a constant, therefore we can solve for n.
Firstly however, if we are going to use P in pascals, T in k, R in J/mole*k, and n in moles (units), then we need to have the volume in cubic meters. Therefore:
V = 284 mL = 284 cm cubed
284 cm cubed * 10 to the negative 6 meters cubed per cm cubed = 0.000284 meters cubed.
Now we can rearrange the gas law to solve for n,
n = PV / RT = (100,000*0.000284) / (8.314510*273.15) = 0.0125 moles
note: do not worry about where the 8.314510 came from it is just a constant.
Because we know that the reaction produced 0.0125 moles of CO2, we know that there must have been this amount of calcite in the original sample. Therefore if we multiply this number by the molecular weight of calcite we will get the number of grams of calcite in the original sample - i.e.:
0.0125 moles * 100.09 g / mole = 1.25 g
Therefore we had 1.25 g CaCO3 in our original sample. Therefore the mass percentage is given by:
MP% = mass CaCO3 / mass sample * 100%
= 1.25 g / 1.506 g * 100% = 83.11% (by mass)
- BBLv 79 years ago
284 mL = 0.284 L CO2. At STP 1 mol occupies 22.4 L. 0.284 / 22.4 = 0.01268 mol CO2.
CO2 / CaCO3 ratio is 1 / 1.
0.01268 mol CaCO3 * mol mass CaCO3 = 0.01268 mol * 100 g/mol = 1.268 g CaCO3.
CaCO3 mass / limestone mass * 100 = 1.268 / 1.506 * 100 = 84.2% CaCO3.
- 9 years ago
first you balance the equation:
CaCO3------->CaO + CO2
they all react in a 1 to 1 mole ratio
since 284 mL of teh gas was produced at STP, you determine the number of moles of CO2
1 mole CO2 = 22.4dm3
x moles CO2=0.284dm3
hence 0.013 moles of calcium carbonate was in teh original sample, to find the mass of calcium carbonate
0.013 * 100=.1.3grams
1.3/1.506 * 100= 86.32%
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- 9 years ago
since 22.4 l ( or dm3) of any gas equals to one mole of the gas under STP.. we can calculate the number of moles of CO2 = 0.012 moles..
from the equation we can see the mole ration between CO2 and CaCO3 is 1:1..
hence number of moles of CaCO3 is also 0.012..
from the molecular mass of calcium carbonate.. we can get the mass of caco3
since 1 moles has the mass of 100.1 ( ca + c + 3O)
0.012 has the mass of 1.2012..
1.2012 / 1.506 x 100 = 79.8% :) :) Hope I helped..