Question

If a utility burned 7.57 million tons of coal that was 2.00% sulfur by weight, how many tons of sulfur dioxide were emitted? Answer in scientific notation.

Answer 1

Step-by-step explanation:

2% of 2million tons of sulphur=

2÷100×7.57E9

That is;

`(2)/(100) * 7.57 * 10 {}^(6) = 0.02 * 7.57 * 10 {}^(6) \\ = 2 * 10 {}^( - 2) *7.57 * 10 {}^(6) = 2 * 7.57(10 {}^( - 2 + 6)) \\ = 15.14 * 10 {}^(4) (for \: sulphur) \\ for \: sulphur \: dioxide = so2 \\ the \: molar \: mass \: of \: so2 = 64g.mol {}^( -1) \\ if \: 32grams \: of \: sulphur \: weighs \: 15.14 * 10 {}^(4) tones \\ 64grams \: will \: weigh \: (64)/(32) * 15.14 * 10 {}^(4) \\ = 2 * 15.14 * 10 {}^(4) \\ = 30.28 * 10 {}^(4) \\ or = 3.028 * 10 {}^(5) tones`

Answer 2

So mass of sulphur

`\\ \rm\Rrightarrow 0.02* 757* 10^(6)`

`\\ \rm\Rrightarrow 1514* 10^4`

• S+O_2–≥SO_2

Moles =64/32=2

So

So tons of sulphur dioxide

`\\ \rm\Rrightarrow 2(1516)10^4`

`\\ \rm\Rrightarrow 3032* 10^4ton`