Question

1) A 2.00 g sample of gas occupies 8.40 m3 at 0°C and 101.3 kPa pressure. Calculate the volume at 0°C and a pressure of 84.0 kPa.

2) Small amounts of hydrogen are conveniently prepared by reacting zinc with hydrochloric acid. Zn + 2HCl --> ZnCl2 + H2 How many grams of zinc are required to prepare 2.50 L H2(g) at 765 Torr and 22°C?

Answer 1

1)           Initial conditions                                    Final conditions

               P1= 101.3Kpa                                        P2= 84.0 Kpa

               V1 = 8.40m3                                           V2 =

               According to Boyles law, at constant temperature and given mass of a gas

P1V1= P2V2

101.3 x 8.4 = 84.0 x V2

V2= 10.13 m3

Volume of gas = 10.13 m3

2)    Zn + 2 HCl ------------------- ZnCl2 + H2

P= 765 torr = 1.00658 atm

V= 2.5L

T= 22C = 22+273= 295K

R= 0.0821 L-artm/mole-K

PV= nRT

n= PV/RT = 1.00658 x 2.5/0.0821x295= 0.104 moles

number of moles of H2= 0.104 moles

According to equation

1 mole of H2 = 1mole of Zn

0.104 mole of H2 = 0.104 moles of Zn

number of moles of Zn= 0.104 mole

molar mass of Zn= 65.38 gram/mole

mass of 0.104 mole of Zn = 0.104 x 65.38= 6.799 grams

mass of Zn= 6.8 grams.