Question

How many grams of co2 are contained in 550 ml of the gas at stp?

Answer 1

v = 550 mL = 0.550 L
V = 22.4 L/mol   (STP)
M(CO₂)=44.01 g/mol

n(CO₂)=v/V

m(CO₂)=n(CO₂)M(CO₂)=vM(CO₂)/V

m(CO₂)=0.550*44.01/22.4=1.08 g

Answer 2

There are 1.08 grams of co2 contained in 550 ml of the gas at stp

Given that    Volume of gas   =550 ml   = 0. 550  L

                 Standard Temperature  =  T  = 273.15 K

                 Standard Pressure  =  P  =  1 atm

                 Gas Constant  =  R  =  0.08205 L⋅atm⋅K⁻¹⋅mol⁻¹

        Mass=?

From the question,    the CO₂ gas is acting ideally so we will apply Ideal Gas equation;

                              P V  =  n R T

Solving for n,

                             n  =  P V / R T

Putting values,

               n  =  1 atm × 0.55 L / 0.08205 L⋅atm⋅K⁻¹⋅mol⁻¹ × 273.15 K

               n  =  0.0245 moles

Remember that  

Number of Moles  =  Mass / Molar.Mass

Solving for Mass,

Mass  =  Moles × M.Mass

Mass  =0.0245 mol × 44.01 g/mol

Mass = 1.08 grams of CO₂

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