Question

Carbon reacts w sulfur 2 form Carbon disulfide ?

Write a balanced chemical equation and answer the questions below.

How many moles of product are produced when 9.3 moles of sulfur are used?
How many moles of product are produced when 22 grams of Carbon are used?

Answer 1

This reaction is pretty antiquated, in industry nowadays carbon gas derivatives are used for thermodynamic reasons. This could occur during volcanic eruptions, though.

`C(s) + 2S(s) to CS_2(s)`

From here on it's simple stoichiometry! Think of it as a puzzle of equalities.

`9.3"mol" * (CS_2)/(2S) approx 4.7"mol"` of carbon disulfide, and

`22"g" * "mol"/(12"g") * (CS_2)/C approx 1.8"mol"` of carbon disulfide

are produced given your reagent specifications.

Answer 2

`"9.3 mol S"` will produce `"4.7 mol CS"_2"`.

`"22 g C"` will produce `"1.8 mol CS"_2"`.

Explanation:

Balanced equation

`"C(s) + 2S(s) +"Delta"` `rarr` `"CS"_2(l)`

Moles `"CS"_2"` from `"9.3 mol S"`

Multiply mol `"S"` by the mol ratio between `"S"` and `"CS"_2"` in the balanced equation, with `"CS"_2"` in the numerator.

`9.3color(red)cancel(color(black)("mol S"))xx(1"mol CS"_2)/(2color(red)cancel(color(black)("mol S")))="4.7 mol CS"_2"` (rounded to two significant figures)

Moles `"CS"_2"` from `"22 g C"`

Determine moles `"C"` by dividing the given mass by its molar mass `("12.011 g/mol")`. Do this by multiplying the given mass by the inverse of the molar mass (mol/g).

Then multiply by the mol ratio between `"C"` and `"CS"_2"` in the balanced equation, with `"CS"_2"` in the numerator.

`22color(red)cancel(color(black)("g C"))xx(1color(red)cancel(color(black)("mol C")))/(12.011color(red)cancel(color(black)("g C")))xx(1"mol CS"_2)/(1color(red)cancel(color(black)("mol C")))="1.8 mol CS"_2"` (rounded to two significant figures)