In this problem, assume that all gas volumes are measured at a temperature 320 K

Question

In this problem, assume that all gas volumes are measured at a temperature 320 K and pressure 120 kPa (i.e., the product(s) of the reaction as well as the reactants). The gases H_2 and O_2 react to form water vapor, H_2O. The chemical equation that describes this reaction is 2 H_2 + O_2 rightarrow H_2O. This can be interpreted as follows: Two molecules of hydrogen (H_2) react with one oxygen (O_2) molecule to form two molecules of water (H_2O). Suppose that 600 ml of H_2 take part in a reaction with O_2. Find an expression for the number of moles in the hydrogen sample. How many hydrogen molecules are in the sample? How many moles of O_2 are needed to react with this amount of hydrogen? What is the volume of O_2 needed? (You should be able to find a numerical answer.) How many moles of H_2O are produced in this reaction? What is the volume of the H_2O produced? If the reaction were carried out again at a different (constant) pressure and temperature, would the ratio of volumes of hydrogen and water change? Explain your reasoning. Do your answers to part C depend on the relative size or the relative mass of the reactant particles? (For example, would your answer change if the reaction were between N_2 and O_2 to produce nitrous gas, N_2O?) Explain your reasoning.

Explanation / Answer

A) Moles of H2 :

Given that T = 320 K

P = 120 kPa = 120 x 0.00987 atm = 1.184 atm

volume of H2 , V= 600 mL = 0.6 L

moles of H2, n = ?

Ideal gas equation is PV = nRT where R = Universal gas constant = 0.0821 L.am/K/mol

n = PV/RT

= (1.184 atm x 0.6 L) / (0.0821 L.am/K/mol x 320 K)

= 0.027 moles

Therefore,

moles of H2 = 0.027 mol

Molecules of Hydrogen :

1 mole of any smaple contains Avogadro's number of molecules i.e. 6.022 x 1023 molecules

Therefore,

Molecules of H2 = 0.027 x 6.022 x 1023 molecules

= 0.16 x 1023 molecules

B) Moles of O2:

2H2 + O2 ----------> 2H2O

2mol 1 mol   

0.027 mol ?

Then,

? = ( 0.027 mol H2/ 2 mol H2 ) x 1 mol O2

= 0.0135 mol O2

Therefore,

0.0135 moles of O2 are required.

Volume of O2 :

Given that T = 320 K

P = 120 kPa = 120 x 0.00987 atm = 1.184 atm

volume of O2 , V = ? L

moles of O2, n = 0.0135 mol

Ideal gas equation is PV = nRT where R = Universal gas constant = 0.0821 L.am/K/mol

V = nRT/P

= (0.0135 mol x 0.0821 L.am/K/mol x 320 K)/ 1.184 atm

= 300 mL

Therefore,

300 mL of O2 are required.

  

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