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In a electricity producing plant $3.1\times10^7\ \mathrm{kg}$ coal with $2.4\ \%$ rhombic sulfur is burnt every year. What is the volume of the produced $\ce{SO2}$ gas at STP?

My effort for solving this answer :

In one day how much sulfur is burnt. Then I wrote the equation

$$\ce{S + O2 -> SO2}$$

With the unitary method I tried but I am confused from where to start, how to start? Please give me a hint how to start the answer.

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closed as off-topic by Mithoron, airhuff, Todd Minehardt, Jon Custer, Tyberius Mar 15 '18 at 22:36

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You have an excess of oxygen, right...?

So the limiting reactant is the sulfur, which is $2.4$% of the coal.

Spoiler: Step 1:

So calculate the sulfur content, which is $3.1 \times 10^7$ kilograms multiplied by $2.4 \times 10^{-2}$ which is $7.44 \times 10^5$ kilograms of sulfur.

Step 2:

The $\ce{SO_2}$ is $50$% sulfur by mass (32 amu of sulfur vs. 64 amu total) so there is $1.488 \times 10^6$ kilograms of $\ce{SO_2}$ produced.

Step 3 (for moles, optional):

If you want moles, divide by $64$, which is $2.325 \times 10^4$ moles (someone check my math)

Then again, you emphasized STP, so proceed with caution (sulfur doesn't react at $20$ degrees Celsius).

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  • $\begingroup$ Thanks , I have started to think your hint , soon I will get my answer :) $\endgroup$ – user70421 Mar 15 '18 at 18:37

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