Question

11.4 gm of a mixture of butene, c4h8 and butane c4h10, was burned in excess oxygen. 35.2 gm of c02 and 16.2 gm of h20 were obtained. Calculate the percentage by mass of butane in original mixture.

Answer 1

Answer:

combustion reaction will look like this:

Fuel + Oxygen ---(ignition)--> Carbon Dioxide + Water

In this case, the fuels are butane (C4H10) and butene (C4H8). So we'll do an equation for each:

2 C4H10 + 13 O2 --> 8 CO2 + 10 H2O

C4H8 + 6 O2 --> 4 CO2 + 4 H2O

Now let's calculate how much CO2 we'd get if the mixture were 100% butane vs 100% butene:

2.86 g butane / 58.12 g mol−1 = 0.04921 mol butane

0.04921 mol butane * 8 mol CO2 / 2 mol butane = 0.1968 mol CO2 * 44.009 g/mol = 8.662 g CO2

2.86 g butene / 56.11 g mol−1 = 0.05097 mol butene

0.05097 mol butene * 4 mol CO2 / 1 mol butene = 0.2039 mol CO2 * 44.009 g/mol = 8.973 g CO2

Let x% be the percent butane, and y% be the percent butene.

x% * 8.662 g + y% * 8.973 g = 8.80 g

y% = 100% - x%

x% * 8.662 g + (100% - x%) 8.973 g = 8.80 g

8.662 g * x% - 8.973 g * x% + 8.973 g = 8.80 g

-0.31031 g * x% = -0.1728 g

x = 55.7%

y = 44.3%