Question

Co (5.00 g) and co2 (5.00 g) were placed in a 750.0 ml container at 50.0°c. the total pressure in the container was __________ atm.

Answer 1

10.33 atm

Further explanation

Given:  

CO (5.00 g) and CO₂ (5.00 g) were placed in a 750.0 ml container at 50.0°C.

Question:  

The total pressure in the container was ... atm.

The Process:

Step-1: preparing for the molar mass of gases (Mr)

• Mr CO = 12 + 16 = 28 g/mol
• Mr CO₂ = 12 + 2(16) = 44 g/mol

Step-2: find out the number of moles of gases

Mole conversions $\boxed{ \ moles = \frac{mass}{Mr} \ }$

• Moles of CO = $\boxed{ \ moles = 5.00 \ g \times \frac{1 \ mol}{28 \ g} \ } \rightarrow \boxed{ \ = 0.17857 \ moles \ }$
• Moles of CO₂ = $\boxed{ \ moles = 5.00 \ g \times \frac{1 \ mol}{44 \ g} \ } \rightarrow \boxed{ \ = 0.11364 \ moles \ }$

Therefore the number of moles of gases equal to 0.17857 + 0.11364 = 0.29221 moles.

Step-3: find out the total pressure in the container

We use an equation of state for an ideal gas:  

$\boxed{\boxed{ \ \frac{pV}{T} = nRT \ }}$  

• p = pressure (in atm)  
• V = volume (in L) , i.e., 750.0 ml = 0.75 L
• n = moles
• R = 0.0821 atm•L•mol⁻¹•K⁻¹ as the molar gas constant
• T = temperature (in Kelvin) , hence 50°C + 273 = 323 K

Prepare p as the subject you want to find.

$\boxed{ \ p = \frac{nRT}{V} \ }$

$\boxed{ \ p = \frac{0.29221 \times 0.0821 \times 323}{0.75} \ }$

Thus the total pressure in the container was 10.33 atm.

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Learn more  

1. To what temperature would you need to heat the gas to double its pressure? brainly.com/question/1615346#  
2. Find out the partial pressure in the container brainly.com/question/7141116
3. Find out he molecular weight of a gas that has a density of 5.75 g/L at STP brainly.com/question/7497852

Answer 2

The molar mass of CO is 28 g/mol while that of CO2 is 44 g/mol. Let us calculate the total moles present in the container.

total moles = [5g / (28 g/mol)] + [5g / (44 g/mol)]

total moles = 0.2922 mol

 

Using PV = nRT, we get the pressure:

P = nRT / V

P = (0.2922 mol * 0.0821 L atm/mol K * 323.15 K) / 0.75 L

P = 10.34 atm