Question

1. Exercise 11.79

A gas mixture contains 74% nitrogen and 26% oxygen.

If the total pressure is 1.18 atm what are the partial pressures of each component?

Express your answers using two significant figures. Enter your answers numerically separated by a comma.

2. Chapter 11 Question 11 - Algorithmic

What is the molecular weight of a gas if a 21.0 g sample has a pressure of 836 mm Hg at 25.0°C in a 2.00 L flask? (R= 0.0821 L atm/ mol K)

What is the molecular weight of a gas if a 21.0 g sample has a pressure of 836 mm Hg at 25.0°C in a 2.00 L flask? (R= 0.0821 L atm/ mol K)

243 amu

234 amu

1.89 amu

11.1 amu

none of the above

3. Exercise 11.91

CH3OH can be synthesized by the reaction:
CO(g)+2H2(g)→CH3OH(g)

Part A

How many liters of H2 gas, measured at 748 mmHg and 83 ∘C, are required to synthesize 0.53 molof CH3OH?

VH2 =	??????	L

Part B

How many liters of CO gas, measured under the same conditions, are required?

Express your answer using two significant figures.

VCO =	?????	L

4. Exercise 11.97

How many grams of calcium are consumed when 160.4 mL of oxygen gas, measured at STP, reacts with calcium according to the following reaction?
2Ca(s)+O2(g)→2CaO(s)

mCa =		g

Answer 1

1. Given Total Pressure=1.18 atm, 74% N2=0.74, 26% O2=0.26,

From Daltons law, Partial pressure Pgas=X*Ptotal

P_N2= 0.74*1.18=0.8732

P_O2= 0.26*1.18=0.3068

2. From Ideal gas,

PV=nRT

Given, Weight=21g, P=836mmof Hg=836/760=1.1 atm, V=2 L, T=25 degC=298K, R=0.0821 L atm/mol/K

n=PV/RT=(1.1*2)/(0.0821*298)=0.0899,

wt/Mwt=0.0899,

Molecular weigth=0.0899*21=233.592g/mol

3. Part A:

CO+2H2------->CH3OH

Given, P=748 mm of Hg=748/760=0.9842 atm, T=83 deg C=356 K, R=0.0821 L atm/mol/K, n_CH3OH=0.53mol

moles of H2=2*moles of CH3OH=0.53*2=1.06mol

from Ideal gas equation, PV=nRT,

V_H2=nRT/P=(1.06*0.0821*356)/0.9842=31.4786 L

V_H2= 31.4786 L

3. Part B:

The volume of CO gas under the same conditions is half the volume of H2 gas, because 1mole of CH3OH come from 1mole of CO and 2 moles of H2

V_H2=nRT/P=(0.53*0.0821*356)/0.9842=31.4786 L

Therefore V_CO=15.7393 L

4. Given at STP, 1 mole of ideal gas=22.4L, volume of O2=160.4ml=0.1604 L

moles of O2=0.1604/22.4=0.00716 mol

Freom the reaction, 2Ca+O2-------->2CaO

moles of Ca=2*moles of O2=0.01432mol

n=Wt/MW, Weight=n*MW=0.5728 g

grams of calcium=0.5728 g