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

So mass of sulphur

$\\ \rm\Rrightarrow 0.02\times 757\times 10^{6}$

$\\ \rm\Rrightarrow 1514\times 10^4$

• S+O_2–≥SO_2

Moles =64/32=2

So

So tons of sulphur dioxide

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

$\\ \rm\Rrightarrow 3032\times 10^4ton$

Answer 2

Explanation:

2% of 2million tons of sulphur=

2÷100×7.57E9

That is;

$\frac{2}{100} \times 7.57 \times 10 {}^{6} = 0.02 \times 7.57 \times 10 {}^{6} \\ = 2 \times 10 {}^{ - 2} \times7.57 \times 10 {}^{6} = 2 \times 7.57(10 {}^{ - 2 + 6}) \\ = 15.14 \times 10 {}^{4} (for \: sulphur) \\ for \: sulphur \: dioxide = so2 \\ the \: molar \: mass \: of \: so2 = 64g.mol {}^{ -1} \\ if \: 32grams \: of \: sulphur \: weighs \: 15.14 \times 10 {}^{4} tones \\ 64grams \: will \: weigh \: \frac{64}{32} \times 15.14 \times 10 {}^{4} \\ = 2 \times 15.14 \times 10 {}^{4} \\ = 30.28 \times 10 {}^{4} \\ or = 3.028 \times 10 {}^{5} tones$