How to solve limiting reagent problem set?

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Can someone please explain to me how to do question 1 and 2? Thanks!

1 Answer
Dec 8, 2017

3.4xx"10"^10" g O"_2" is required for one day's coal production.

4.7xx"10"^10" g CO"_2" is produced each day.

All answers are rounded to two significant figures.

Explanation:

The basic process will be to first determine the moles and then determine the mass.

"mol A"="mass A"xx("mol A")/("molar mass A")
color(white)(.......................................)uarr
color(white)(..........................) (Look up or calculate).

Convert mol A to mol B using the mol A: mol B ratio from the balanced equation.

"mol B"="mol A"xx("mol B")/("mol A")

Then determine the mass from the moles.

"mass B"="moles B"xx("molar mass B")/("mol B")

1. How many grams of oxygen are required for the combustion of 1 day's coal?

First convert tons to grams using dimensional analysis.

"1 ton"="907184.7 g " larr Look this up on the internet.

Conversion factor to use: "907184.7g"/"1 ton".

14000color(red)cancel(color(black)("ton C"))xx(907184.7"g C")/(1color(red)cancel(color(black)("ton C")))=1.27xx"10"^10" g C"

Now that you know the mass of "C" in grams, you can answer the rest of the question.

1.27xx10^10color(red)cancel(color(black)("g C"))xx(1"mol C")/(12.011color(red)cancel(color(black)("g C")))=1.06xx"10"^9" mol C"

1.06xx10^9color(red)cancel(color(black)("mol C"))xx(1"mol O"_2)/(1color(red)cancel(color(black)("mol C")))=1.06xx"10"^9" mol O"_2"

1.06xx10^9color(red)cancel(color(black)("mol O"_2))xx(31.998"g O"_2)/(1color(red)cancel(color(black)("mol O"_2)))=color(blue)(3.4xx"10"^10" g O"_2"

2. How much carbon dioxide is produced each day?

1.06xx10^9color(red)cancel(color(black)("mol C"))xx(1"mol CO"_2)/(1color(red)cancel(color(black)("mol C")))=1.06xx"10"^9" mol CO"_2"

1.06xx10^9color(red)cancel(color(black)("mol CO"_2))xx(44.009"g CO"_2)/(1color(red)cancel(color(black)("mol CO"_2)))=color(blue)(4.7xx"10"^10" g CO"_2"