Question

A mixture of solid calcium oxide, CaO, and solid barium oxide, BaO, weighing 3.458 g is added to a 2.00 L flask containing carbon dioxide, CO2, at 30 ∘C. The CaO and BaO react completely with the carbon dioxide gas to form solid CaCO3 and BaCO3, respectively. Calculate the moles of CO2 consumed by the reaction if the initial pressure of CO2 gas was 0.628 atm and the final pressure of CO2 gas is 0.232 atm.

moles of CO2:

What is the mass percentage of CaO in the solid mixture?
mass percent CaO: %

What is the mass percentage of BaO in the solid mixture?
mass percent BaO: %

Can someone help me get the answers to these question Please!! I am very confused, and is due in couple of hours :'(

Answer 1

Answer:

no idea

Explanation:

Answer 2

Answer:

The mass percentage of CaO in the solid mixture = = 22.55%.

The mass percentage of BaO in the solid mixture = = 77.21%. 

Explanation:

Given,

Weight of mixture of solid calcium oxide, CaO, and solid barium oxide, BaO = 3.458g

The volume of carbon dioxide, CO₂, V = 2L

Temperature, T = 30°C

Convert °C into K

• 1°C = 273.15 K
• 30°C = (30 + 273.15) K = 303.15K

The initial pressure of carbon dioxide, CO₂, $P_{i}$ = 0.628 atm

The final pressure of carbon dioxide, CO₂, $P_{f}$ = 0.232 atm

The number of moles of carbon dioxide, CO₂, n =?

Now, we can find out the number of moles of CO₂ before the reaction, by using the ideal gas equation given below:

• $n_{i} = \frac{P_{i}V }{RT}$

Here,

• n = The number of moles of CO₂
• $n_{i}$ = The initial pressure of CO₂
• V = The volume of CO₂
• R = The gas constant = 0.0821 Latm/Kmol
• T = Temperature

After putting all the values in the ideal gas equation, we get:

• $n_{i} = \frac{0.628\times2 }{0.0821\times 303.15}$
• $n_{i} = \frac{1.256 }{24.88}$ = 0.05 mol

Now, the number of moles of CO₂ after the reaction:

• $n_{f} = \frac{P_{f}V }{RT}$
• $n_{i} = \frac{0.232\times2 }{0.0821\times 303.15}$
• $n_{i} = \frac{0.464 }{24.88}$ = 0.0186 mol

Therefore,

The amount of CO₂ that is consumed by the reaction:  

• $n = n_{i} - n_{f}$
• n = 0.05 - 0.0186 = 0.0314 mol

As we know, CO₂ and CaO react in the same mole ratio, i.e., 1:1.

Also, the mole ratio of CO₂ and BaO is the same.

Therefore,

• Mole of CO₂ + Mole of BaO = 0.0314 mol

Let the mass of CaO = x g

Then the mass of BaO = (3.458 - X) g

As we know,

• Molar mass of CaO = 56.08g/mol
• Molar mass of BaO = 153.3g/mol

Now,

• The number of moles can be calculated by the formula given below:
• $n =\frac{mass}{molar mass}$

Therefore,

• Number of moles of CaO = $\frac{x}{56.08}$ mol
• Number of moles of BaO= $\frac{3.458 - x}{153.3}$ mol

As we know,

• Mole of CO₂ + Mole of BaO = 0.0314 mol
• $\frac{x}{56.08} + \frac{3.458 - x}{153.3} = 0.0314$
• $\frac{153.3x + 193.9 - 56.08x}{8597.06} = 0.0314$
• 97.22x + 193.9 = 269.9
• 97.22x = 76
• x = 0.78 g

Therefore,

• The mass of CaO = x g = 0.78g
• The mass of BaO = (3.458 - X) g = 3.458 - 0.78 = 2.67g

Now,

• The mass percentage of CaO in the solid mixture = $\frac{0.78}{3.458} \times 100$ = 22.55%.
• The mass percentage of BaO in the solid mixture = $\frac{2.67}{3.458} \times 100$ = 77.21%.