How many grams of CO2,(g) would be found in a 5.6 L container at 23oC that has a pressure of 760 mm Hg?

Mass = 10.12 g

Explanation:

Given data:

Volume of CO₂ = 5.6 L

Pressure = 760 mmHg

Temperature = 23°C

Mass of CO₂ = ?

Solution:

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

Now we will convert the temperature.

23+273 = 296 K

760/760 = 1 atm

1 atm × 5.6 L = n × 0.0821 atm.L/ mol.K  × 296K

5.6 atm.L =  n × 24.30 atm.L/ mol

n = 5.6 atm.L  / 24.30 atm.L/ mol

n = 0.23 mol

Mass of CO₂:

Mass = number of moles × molar mass

Mass = 0.23 mol × 44 g/mol

Mass = 10.12 g

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g