1. Exercise 11.79 A gas mixture contains 74% nitrogen and 26% oxygen. If the tot

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)

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?

  L

Part B

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

Express your answer using two significant figures.

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)

Explanation / Answer

Answers:

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

From Daltons law, Partial pressure Pgas=X*Ptotal

PN2= 0.74*1.18=0.8732

PO2= 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, nCH3OH=0.53mol

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

from Ideal gas equation, PV=nRT,

VH2=nRT/P=(1.06*0.0821*356)/0.9842=31.4786 L

VH2= 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

VH2=nRT/P=(0.53*0.0821*356)/0.9842=31.4786 L

Therefore VCO=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

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