What volume of CO 2 (g) will be formed when 2.0 g CaCO 3 (s) is heated and the g

Question

What volume of CO2(g) will be formed when 2.0 g CaCO3(s) is heated and the gas is collected at 700 mm pressure and 20 oC? The reaction is: CaCO3(s) CaO(s) + CO2(g)

A. 0.44 L                     B. 0.52 L                     C. 52 L                                    D. 3.61 L                     E. 22.4 L

What is the mass of CO2 in a 340. mL container when its pressure is 2.80 atm and the temperature is 57 oC?

A. 26.7 g                     B. 0.125 g                    C. 125 g                      D. 3.52 g                     E. 3.47x10–4 g

Explanation / Answer

CaCO3(s) CaO(s) + CO2(g)

The equation tells us that 1 mole of CO2 will be produced by 1 mole of CaCO3(s) ( a 1:1 ratio).

Therefore, moles of CO2 = moles of CaCO3(s)

Calculate the moles of CaCO3(s) from the starting mass of the CaCO3(s):

moles CaCO3   = 2.0 g / 100.08 g/mol= 0.0199 mol CaCO3   = 0.0199 mol CO2

Then convert the moles of CO2 to mass:

                 mass of CO2   =   0.0199 moles x   44 g/mol    = 0.879 gm

0.879 gm of CO2(g) will be formed

Density of CO2 is 1.98 gm/L

Volume of CO2(g) = 0.879 gm / 1.98 gm/L = 0.44 L

0.44 L of CO2(g) will be formed

question 2)

Solution:

Given

temperature is 57 oC = 273 + 57 = 330 K

volume = 340. mL = 0.340 L

Ideal Gas Law PV = nRT

Rearrange the Ideal Gas Law to this:

n = PV / RT

Substitute values into the equation:

n = [ (2.80 atm) (0.340 L) ] / [ (0.08206 L atm mol¯1 K¯1) (330.0 K) ]

n = 0.035 moles

Then convert the moles of CO2 to mass:

                 mass of CO2   =   0.035 moles x   44 g/mol    = 1.54 gm

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