14.A 4.25 gram sample of an unknown gas is found to occupy a volume of 1.70 L at a pressure of 883 mm Hg and a temperature of 58 °C. The molar mass of the unknown gas is g/mol.

58.5g/mol

Explanation:

Given parameters:

Mass of gas  = 4.25g

Volume  = 1.7L

Pressure  = 883mmHg

                760mmHg  = 1 atm

                883mmHg  =   = 1.16atm

Temperature  = 58°C  = 58 + 273  = 331K

Unknown:

Molar mass of sample  = ?

Solution:

To solve this problem, we use the ideal gas equation to find the number of moles;

            PV  = nRT

 P is the pressure

V is the volume

n is the number of moles

R is the gas constant  = 0.082atmdm³mol⁻¹K⁻¹

 T is the temperature

           n  =   =   = 0.07mole

 Since;

            mass  = number of moles x molar mass

            4.25  = 0.07 x molar mass

 Molar mass  = 58.5g/mol

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

it would be a gas formation