Question

How to solve limiting reagent problem set?

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

Answer 1

`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"`