Question

Part A

A common way to make hydrogen gas in the laboratory is to place a metal such as zinc in hydrochloric acid (see the figure). The hydrochloric acid reacts with the metal to produce hydrogen gas, which is then collected over water. Suppose a student carries out this reaction and collects a total of 145.8 mL of gas at a pressure of 749 mmHg and a temperature of 25∘C. What mass of hydrogen gas (in milligrams) does the student collect? (The vapor pressure of water is 23.78 mmHg at 25∘C.)

Part B)

Consider the following chemical reaction:

2H2O(l)→2H2(g)+O2(g)

What mass of H2O is required to form 1.6 L of O2 at a temperature of 320 K and a pressure of 0.987 atm ?

Express your answer using two significant figures.

Part C)

What volume of O2 gas, measured at 760 mmHg and 17 ∘C, is required to completely react with 50.8 g of Al?

Express the volume in liters to three significant figures.

Answer 1

Part A :

Let us see what are the given parameter

Volume = V =145.8 mL = 0.1458 L

The pressure of gas is due to hydrogen gas and water vapours

so pressure = Ptotal = PH2 + P H2O vapours

749 = PH2 + 23.78

P_H2 = 725.22 mmHg

Also, 760 mmHg = 1 atm

So 725.22mmHg = 0.954 atm

To calculate mass we will first calculate moles of H2 by using ideal gas equation

PV = nRT

n = moles = PV / RT

R = 0.0821

T = 298 K
moles = 0.954 X 0.1458 / 298 X 0.0821

Moles of hydrogen = 0.00568

Mass of hydrogen = Moles X molecular weight = 0.00568 X 2 = 0.01136 grams
1 gram = 1000 mg

0.01136 grams = 11.36mg
B) Let us calculate the moles of O2 formed during reaction by using ideal gas equation , which is

PV=nRT

P = 0.987 , V = 1.6 L   T = 320 K   R = 0.0821

n = moles = 0.0601

Now as per stoichiometry of given reaction

2H2O(l)→2H2(g)+O2(g)

one mole of O2 is formed from 2 moles of water

so for 0.0601 moles of oxygen the moles of water required = 2 X 0.0601 moles = 0.1202 moles

Mass of 1 mole of water = 18 grams

so mass of 0.1202 moles of water = 18 X 0.1202 grams = 2.1636 grams

C) The reaction of Al with O2 is

2Al + 3/2O2 --> Al2O3

so as per stoichiometry of equation, 2 moles of Al will react with 3/2 moles of O2

Atomic weight of Al = 27

Moles of Aluminium reacted = 50.8 grams / 27 = 1.881 moles

so moles of oxygen required = 3 X 1.881 /4 = 1.4107 moles

Volume can be calculated from ideal gas equation

PV = nRT

Volume = moles X R X temperature / Pressure

Volume = 1.4107 X 0.0821 X 290 / 1 atm

Volume = 33.58 Litres