Question

A sample of gas occupies 300 ml at STP. Under what pressure would this sample occupy a volume of 150 ml at a temperature of 546 ̊C?

Calculate the pressure exerted by 60.0 g C2H6 (ethane) in a 30.0-L vessel at 27 ̊C.

When nickel is heated in carbon monoxide, a gaseous product is formed that has a density of 6.57 g/L at 750 torr and 40 ̊C. The compound is analyzed to show 34.30% Ni, 28.1% C and 37.53% O. What is the molecular formula of the compound?

If 2.0 liters of N2 at 2.0 atm and 2.0 liters of H2 at 1.0 atm are put in a 4.0-liter vessel, what is the final pressure, temperature being constant?

Two gases, Ar and He, are contained in vessel. The mole fraction of He is 0.35 with a partial pressure of 1.15 atm. What is the partial pressure of Ar?

Given the reaction below, how many grams of iron are needed to produce 100 liter of hydrogen gas at STP?

3Fe(s) + 4H2O(l) → Fe3O4 (s) +4H2 (g)

What volume of oxygen at 18 ̊C and 750 torr can be obtained from the thermal decomposition of 100-g KClO3?

2KClO3 (s) → 2KCl(s) +3O2 (g)

A student collects 2.63 L of O2 over water at 29 ̊C and 769 torr by thermal decomposition of potassium chlorate. a) How many grams of oxygen are in the sample? b) How many grams of KClO3 decomposed to produce the oxygen?

2KClO3 (s) → 2KCl (s)+ 3O2 (g)

Answer 1

Answer – 1) We are given, volume, V1 = 300 mL, P1 = 1.00 atm, T1 = 273 K

V2 = 150 mL , T2 = 546 +273 = 819 K

We know, combine gas law

P1V1/T1 = P2V2/T2

So, P2 = P1V1T2 / T1V2

           = 1.00 atm * 300 mL * 819 K / 150 mL * 273 K

           = 6.00 atm

6.00 atm pressure would this sample occupy a volume of 150 ml at a temperature of 546 ̊C

2) Given, mass of ethane , C₂H₆ = 60.0 g , T = 273 +273 = 300 K , V = 30.0 L

First we need to convert the mass to moles

Moles of C₂H₆ = 60.0 g / 30.07 g.mol^-1

                         = 2.0 moles

We know Ideal gas law

PV = nRT

So, P = nRT/V

        = 2.00 moles * 0.0821 L.atm.mol^-1.K^-1 *300 K / 30.0 L

       = 1.64 atm

So, 1.64 atm pressure exerted by 60.0 g C2H6 (ethane) in a 30.0-L vessel at 27 ̊C.

3) We are given, density of gas = 6.50 g/L , P = 750 torr /760 = 0.989 atm

T = 40 +273 = 313 K

So molar mass of gas

Density = PM/RT

So, M = density * RT / P

            = 6.50 g.L^-1* 0.0821 L.atm^-1*313 K / 0.989 atm

            = 169.3 g/mol

Assume, 100 g of sample

So, Ni = 34.30 g , C = 28.1 g , O = 37.53 g

Moles of Ni = 34.30 g / 58.693 g.mol^-1

                    = 0.584 moles

Moles of C = 28.1 g / 12.01 g.mol^-1

                   = 2.34 g

Moles of O = 37.53 g / 15.998 g.mol^-1

                   = 2.34 g

So, moles of Ni is lowest

Ni = 0.584 / 0.584 = 1

C = 2.34 / 0.584 = 4.0

O = 2.34 / 0.584 = 4.0

So empirical formula is NiC₄O₄

Molecular formula = n * empirical formula

n = molecular mass / empirical formula mass

    = 169.3 / 170

   = 1

So molecular formula = 1 * NiC₄O₄

                                                      = NiC₄O₄

4) We are given, V1 = 2.0 liters of N2 , P1 = 2.0 atm, V2 = 4.0 L, P2 = ?

V1 =2.0 liters of H2 , P1 = 1.0 atm , P2 = 4.0-liter

Partial pressure of N₂

P2 = P1V1/V2

      = 2.0 L * 2.0 atm / 4.0L

      = 1.0 atm

Partial pressure of H₂

P2 = P1V1/V2

      = 2.0 L * 1.0 atm / 4.0L

      = 0.5 atm

So total pressure = 1.0 + 0.5 = 1.5 atm