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So2 (5.00 g) and co2 (5.00 g) were placed in a 750.0 ml container at 50.0 °c. the partial pressure of co2 in the container was __________ atm.

Respuesta :

Kalahira
4.02 atm. Since we're looking for just the partial pressure of CO2, we can safely ignore the sulfur dioxide since the total pressure in the container will be the sum of all the partial pressures and the partial pressure will be the pressure that the substance would have if it were the only substance in the container. So, let's determine the number of moles of CO2 we have. Start by looking up the atomic weights: Atomic weight carbon = 12.0107 Atomic weight oxygen = 15.999 Molar mass CO2 = 12.0107 + 2 * 15.999 = 44.0087 g/mol Moles CO2 = 5.00 g / 44.0087 g/mol = 0.113613899 mol The ideal gas law is PV = nRT P = Pressure V = Volume n = number of moles R = Ideal Gas Constant (0.082057338 L*atm/K*mol ) T = Absolute temperature Converting from C to K, gives 50.0 + 273.15 = 323.15 K Take the ideal gas law and solve for P, then substitute known values and calculate result. PV = nRT P = nRT/V P = (0.113613899 mol)(0.082057338 L*atm/K*mol)(323.15 K)/(0.75 L) P = (0.009322854 L*atm/K)(323.15 K)/(0.75 L) P = (3.012680306 L*atm)/(0.75 L) P = 4.016907075 atm Since we only have 3 significant digits of precision available, round to 3 significant digits, getting 4.02 atm.
BladeRunner212

4.02 atm

Further explanation

Given:  

SO₂ (5.00 g) and CO₂ (5.00 g) were placed in a 750.0 ml container at 50.0°C.

Question:  

The partial pressure of CO₂ in the container was ... atm.

The Process:

This problem is only required to find the partial pressure of CO₂ in containers, so we ignore SO₂ in the calculation.

Step-1: preparing for the molar mass of carbon dioxide

Mr CO₂ = 12 + 2(16) = 44 g/mol

Step-2: find out the number of mole of CO₂

Mole conversions [tex]\boxed{ \ moles = \frac{mass}{Mr} \ }[/tex]

Therefore, moles of CO₂ = [tex]\boxed{ \ moles = 5.00 \ g \times \frac{1 \ mol}{44 \ g} \ } \rightarrow \boxed{ \ = 0.11364 \ moles \ }[/tex]

Step-3: find out the partial pressure of CO₂ (in atm)

We use an equation of state for an ideal gas:  

[tex]\boxed{\boxed{ \ \frac{pV}{T} = nRT \ }}[/tex]  

  • 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.

[tex]\boxed{ \ p = \frac{nRT}{V} \ }[/tex]

[tex]\boxed{ \ p = \frac{0.11364 \times 0.0821 \times 323}{0.75} \ }[/tex]

Thus the partial pressure of CO₂ in the container was 4.018 atm, rounded up to 3 significant digits, we get 4.02 atm.

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