1) A sample of methane gas is introduced into an evacuated 5.0 L container, unti

Question

1) A sample of methane gas is introduced into an evacuated 5.0 L container, until a pressure of 3.2 atm is attained. What volume change is required to reduce the methane pressure to 1.0 atm (at constant temperature)?

2) A gas sample stored in a 11.4-L container at 25 °C and 735 torr is compressed until the volume is 7.90 L and the temperature is 72 °C. What is the final pressure inside the container

3) 1.92 g of zinc metal reacts with excess hydrochloric acid to produce hydrogen gas. What volume (L) of hydrogen is generated at STP?

Zn (s) +    2 HCl (aq) ?    ZnCl2  (aq)     +    H2 (g)

Explanation / Answer

1. Volume change is required to reduce the methane pressure to 1.0 atm = 5 L container need to change to 16 L container.

2. Final pressure inside the container = 1228 torr.

3. What volume (L) of hydrogen is generated at STP = 0.6578 L

------------ calculations------------

Applying gas law, P1V1 = P2V2 at constant temperature

P1 = 3.2 atm, V = 5.0 L, P2 = 1.0 atm

V2 = P1V1/P2 = 3.2 atm x 5.0 L / 1 atm = 16 L

Applying gas law, P1V1/T1 = P2V2/T2

P1 = 735 torr, V1 = 11.4 L, T1 = 25 oC = 298 K

P2 = ?, V2 = 7.90 L, T2 = 72 oC = 345 K

P2 = P1 x V1 x T2 / T1 x V2 = 735 torr x 11.4 L x 345 K / 298 K x 7.90 L = 1227.9 = 1228 torr

According to reaction 1 mole of Zn(s) produces 1 mole of H2(g)

At.Wt of Zn = 65.38 g/mol

1.92 g Zn = 1.92 g / 65.38 g/mol = 0.0294 mole

So the amount of H2 produced is 0.0294 mole

Molar volume of a gas is 22.4 L at STP, i.e. 1 mole of any gas required 22.4 L of volume at STP.

So 0.0294 mole of H2 required = 0.0294 x 22.4 L = 0.6578 L

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