Question

carbon monoxide is a gas that is 43% carbon. carbon dioxide only 27% carbon. suppose that the environmental protection agency (epa) analyzes exhaust from cars, a mixture of carbon monoxide and carbon dioxide. if the epa finds that a 1600 mg sample of exhaust gas has 32% carbon, how many mg of the sample were carbon monoxide, and how many were carbon dioxide?

Answer 1

Answer:

500 mg Carbon Monoxide

1100 mg Carbon Dioxide

Step-by-step explanation:

Weight of total sample = 1600 mg

Percentage of Carbon in Carbon Monoxide = 43%

Percentage of Carbon in Carbon Dioxide = 27%

Percentage of carbon in Mixture = 32%

Let amount of Carbon Monoxide = $x$ mg

Amount of Carbon Dioxide in mixture = (1600-$x$) mg

Amount of carbon in $x$ mg of Carbon Monoxide =

$x \times 43\%\\\Rightarrow x \times \frac{43}{100} ...... (1)$

Amount of carbon in (1600-$x$) mg of Carbon Monoxide =

$(1600-x) \times 27\%\\\Rightarrow (1600-x) \times \frac{27}{100}......(2)$

We know that the sample of 1600 mg contains carbon because of Carbon Monoxide and Carbon Dioxide present in it.

Adding (1) and (2) and equating to 32% of 1600 mg:

$x \times \dfrac{43}{100} + (1600-x) \times \dfrac{27}{100} = 1600 \times \dfrac{32}{100}\\\Rightarrow x \times (43-27) = 1600 \times (32-27)\\\Rightarrow 16x = 1600 \times 5\\\Rightarrow x =500$

Amount of Carbon Monoxide in sample,$x$ = 500 mg

Amount of Carbon Dioxide in sample = (1600 - $x$) = (1600 - 500) = 1100 mg

Hence, answer is

500 mg Carbon Monoxide

1100 mg Carbon Dioxide