Hello guys I need help with these questions please, thanks beforehand!! 1)When h

Question

Hello guys I need help with these questions please, thanks beforehand!!

1)When heated, calcium carbonate decomposes to yield calcium oxide and carbon dioxide gas via the reaction
CaCO3(s)---> CaO(s) + CO2(g) How many grams of calcium carbonate are needed to produce 81.0L of carbon dioxide at STP?     Express your answers numerically in grams.


2)  Butane, C4H10, is a component of natural gas that is used as fuel for cigarette lighters. The balanced equation of the complete combustion of butane is 2C4H10(g) + 13O2(g) ---> 8CO2(g) + 10H2O(l). At 1.00atm and 23 Celsius, how many liters of carbon dioxide are formed by the combustion of 2.20g of butane?
Express your answers  numerically in liters.


3)  To identify a diatomic gas (X2), a researcher carried out the following experiment: She weighed an empty 1.00-L bulb,  then filled it with the gas at 1.10atm and 23.0 Celsius and weighed it again. The difference in mass was 1.27g. Identify the gas. Express your answer as a chemical formula.



4)  Three gases (8.00g of methane, CH4, 18.0g of ethane, C2H6, and an unknown amount of propane, C3H8) were added to the same 10.0-L container. At 23.0 Celsius the total pressure in the container is 4.10atm. Calculate the partial pressure of each gas in the container. Express the pressure values numerically in athospheres, separated by commas. enter the partial pressure of methane first, then ethane, then propane.

5)  A gaseous mixture of O2 and N2 contains 33.8% nitrogen by mass. What is the partial pressure of oxygen in the mixture if the total pressure is 325mmHg? Express your answer numerically in millimeters of mercury.


6)  15.0 moles of gas are in a 8.00L tank at 23.2 Celsius. Calculate the difference in pressure between methane and an ideal gas under these conditions. The van der Waals constants for methane are a=2.300L2 x atm/mol^2 and b=0.0430 L/mol.




please help !!! thankkkksssssssss

Explanation / Answer

1) one mole of CO2 at STP = 22.4 L

81.0 L = 3.616 moles

one mole CaCO3 gives one mole CO2, So you need 3.616 moles of CaCO3 for 3.616 moles of CO2

3.616 moles of CaCO3 = 3.616 *100 = 361.6 gms of CaCO3

2) 2.20 g butane = 2.20 / 58 moles of butane = 0.0379 moles

2 moles of butane give 8 moles of CO2, so 0.0379 moles shall give 0.1517 moles of CO2

Applying Ideal gas Equation :

V = nRT / P = 0.1517 *0.0821*296/1 = 3.687 L of CO2

3) We find out moles of gas corresponding to the mass 1.27 g

n = PV/RT = 1.10 * 1 / 0.0821*296 = 0.04526 moles

1.27 g = 0.04526 moles, so 1 mole = 28.05g = X2

This is the molar mass of Nitrogen. So, the unknown gas was nitrogen. N2

4) 8.0g CH4 = 8 / 16 = 0.5 moles of CH4

18.0 g C2H6 = 18/30 = 0.6 moles of ethane

Now, lets calculate partial pressures:

P (CH4) = nRT/V = 0.5*0.0821*296 / 10 = 1.215 atm

P (C2H6) = nRT/V = 0.6 * 0.0821 * 296 / 10 = 1.458 atm

So, P(C3H8) = 4.10 - (1.215+1.458) = 1.427

moles of C3H8 = PV/RT = 1.427*10 / 0.0821*296 = 0.5872 moles

0.5827 moles = 25.83 g of C3H8

5)Assume mass = 100 g

Then mass of N2 = 33.8 g = 33.8 / 28 = 1.20714 moles

mass of O2 = 100- 33.8 = 66.2 g = 66.2/32 = 2.06875 moles

Partial pressure of O2 = mole fraction of O2 * total pressure = [(2.06875)/(2.06875+1.20714)]*325 = 205.24 mm Hg

6) P (ideal) = nRT/V = 15 * 0.0821 * 296.2 / 8 = 45.59atm

Van der waals' eqn:

(P + an^2/V^2) (V-nb) = nRT

P = [nRT/(V-nb) - an^2/V^2 ] = 41.50 atm

So the difference is : 4.081 atm

Hope this helps. Please rate it the best. Thanks.

ASk doubts if you have.

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