Part A A common way to make hydrogen gas in the laboratory is to place a metal s

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 25C. What mass of hydrogen gas (in milligrams) does the student collect? (The vapor pressure of water is 23.78 mmHg at 25C.)

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.

Explanation / Answer

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

PH2 = 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

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