Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

The following reaction is used to generate hydrogen gas in the laboratory and it

ID: 960731 • Letter: T

Question

The following reaction is used to generate hydrogen gas in the laboratory and it is collected over water If 243 mL of gas is collected of 25 degree C an pressure of 745 mm Hg. what mass of hydrogen is produced? A possibly useful table of water vapor pressures is provided below Mg(s) + 2 HCI(aq) rightarrow MgCl_2(aq) + H_2(g) Determine the volume of O_2 (at STP) formed when 50.0 g of KCIO_3 decomposes according to the following reaction. 2 KCIO_3(s) rightarrow 2 KCI(s) + 3 O_2(g) Determine the theoretical yield and the percent yield if 21.8 g of K_2CO_3 is produced from reading 27.9 g KO_2 with 29.0 L of CO_2 (at STP). 4 KO_2(s) + 2 CO_2(g) rightarrow 2 K_2CO_3(s) + 3O_2(g)

Explanation / Answer

23) total pressure = Pressure of H2 + Pressure of H2O

Pressure of H2O = 23.78 (at 25 C)

so pressure of H2 = 745- 23.78 = 721.22 mm Hg

Now we will use ideal gas equation

PV = nRT

n = PV / Rt = moles

R = ideal gas constant = 0.0821 atm L mol^-1 K_1

V = 243 mL = 0.243 L

P = 721.22 mmHg = 0.949 atm ( 1atm = 760mmHg)

T = 25 C =298 K

moles of hydrogen = PV / RT = 0.243 X 0.949 / 298 X 0.0821 = 0.00942 moles

Mass of 1 mole of hydrgen gas = 2 gram

So mass of 0.0942 moles = 0.00942 X 2 g = 0.01884 grams

24) the reaction is

2KClO3 --> 2KCl + 3O2

As per stoichiometry , 1 mole of KClO3 will give (22.4 X 3 L) 67.2 L of oxygen gas on decompositon

[ 1 mole of O2 will occupy 22.4 L]

Molecular wieght of KcLO3 = 122.5

Mass of KClO3 taken = 50 g

So moles of KClO3 = 50 / 122.5 = 0.408

These moles will give = 0.408 X 3 X 22.4 L of O2 = 27.42 L of oxygen

25) As per the stoichiometry, 4 moles of KO2 ( = 71 X 4 grams = 284 grams) will react with 2 moles ( 2X44 = 88 grams of CO2)

Given: moles of KO2 = 27.9 / 71 = 0.393 moles

Moles of CO2 = volume / 22.4 = 1.295 moles

so 0.393 moles will react with 0.393 / 2 = 0.196 moles of CO2 to give 0.196 moles of K2CO3

So mass of K2CO3 should be produced = moles X molecular weight = 0.196 x 138 = 27.048 grams (theoretical yield)

Actual mass produced = 21.8

So % yield = mass produced X 100 / theoretical yield = 21.8 X 100 / 27.048 = 80.597 %