10cm3 of hydrogen was sparked with 10cm3 of oxygen at s.t.p. calculate the volume of unreacted gas

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10cm3 of hydrogen was sparked with 10cm3 of oxygen at s.t.p. calculate the volume of unreacted gas

This one is a little tricky. And here's why I say that.

When it comes to reaction that involve gases kept under the same conditions for pressure and temperature, it's important to realize that the mole ratios that exist between the species involved in the reaction becomes equivalent to the volume ratio.

So, take a look at the balanced chemical equation for your reaction

#color(red)(2)"CO"_text((g]) + "O"_text(2(g]) -> color(blue)(2)"CO"_text(2(g])#

Notice that you have a #color(red)(2):1# mole ratio between carbon monoxide and oxygen gas. This tells you that the reaction will always consume twice as many moles o carbon monoxide than of oxygen gas.

But since we're working with gases under the same conditions for pressure an temperature, you can say the exact same thing about their volumes.

In other words, the reaction will always consume a volume of carbon monoxide that is twice as large as the volume of oxygen gas.

Now, here's where the tricky part comes in. Notice that the problem provides you with equal volumes of #"CO"# and #"O"_2#.

This is a problem because you know that you'd need a volume of #"CO"# that is twice as large as the volume of #"O"_2#.

You an thus conclude that you're dealing with a limiting reagent. More specifically, carbon monoxide will act as a limiting reagent because it will be consumed before all the oxygen gets a chance to react.

So, you can say that the reaction will only consume

#10color(red)(cancel(color(black)("cm"^3"CO"))) * ("1 cm"^3"O"_2)/(color(red)(2)color(red)(cancel(color(black)("cm"^3"CO")))) = "5 cm"^3color(white)(a)"O"_2#

So, #"10 cm"^3# of #"CO"# wil lreact with #"5 cm"^3# of #"O"_2#, leaving #"5 cm"^3# of #"O"_2# in excess.

Now look at the #color(red)(2):color(blue)(2)# volume ratio that exists between #"CO"# and #"CO"_2#. This tells you that for a given volume of #"CO"# that takes part in the reaction, the reaction produces an equal volume of #"CO"_2#.

So, if #"10 cm"^3# of #"CO"# react, you will get #"10 cm"^3# of #"CO"_2# produced.

This means that after the reaction is finished, the reaction vessel will contain

#V_"gas" = overbrace(V_(O_2))^(color(purple)("in excess")) + overbrace(V_(CO_2))^(color(brown)("produced"))#

#V_"gas" = "5 cm"^3 + "1 0cm"^3 = color(green)("15 cm"^3)#

Equation for the reaction:

2H2O + O2 ----> 2H2O

mole ratio: 2 1 2

available vol. 50cm3 20cm3 -

Reaction ratio : 40cm3 20cm3 40cm3

Residual gas : 10cm3 - 40cm3

Residual gas = unreacted hydrogen + formed water(g)

= 10 cm3 + 40 cm3

= 50cm3 

Equation for the reaction:

2H2O + O2 ----> 2H2O

to calculate volume of residual gas,

since 1 mole O2 is 20cm3,

2 moles of H2 will be 20 * 2 = 40cm3.

Meanwhile volume of H2 given is 50cm3 and 40cm3 was used for the sparking(reaction).

Volume of unreacted gas is 50 - 40 = 10cm3.

volume of residual gas = volume of unreacted gas + volume of product gas

10cm3 + 40cm3 = 50cm3.

Chemistry

JAMB 1984


20 cm 3 of hydrogen gas are sparked with 20cm 3 of oxygen gas in an endimoeter at 373k (100 oC) and 1 atmosphere. The resulting mixture is cooled to 298 K (25 oC) and passed over calcium chloride. The volume of the residual gas is

  • A. 40 cm 3
  • B. 20 cm 3
  • C. 30 cm 3
  • D. 10cm 3
  • E. 5cm 3

2H2 + O2 ------2H2O 2V 1V 2V

20cm3 20cm3 -------


20cm3 10cm3 20cm3
Residual gas = 20cm3 - 10cm3 of 02 = 103


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