- #1

- 21

- 0

## Homework Statement

Calculate what mass of carbon you would need to reduce 15.9g of copper(II) oxide to copper by the reaction:

CuO

_{(s)}+ C

_{(s)}[tex]\rightarrow[/tex] Cu

_{(s)}+ CO

_{(s)}

## Homework Equations

n = m / ar

(number of moles = mass / relative atomic mass)

If you know any more, it would be really useful for me though ;) .

## The Attempt at a Solution

Ar(Cu) = 63.5

Ar(O) = 16.0

Ar(C) = 12.0

CuO = 1 / (63.5 + 16.0) = 79.5g

C = 1 / (12.0) = 12.0g

79.5 / 79.5 * 15.9 = 15.9g of CuO

12.0 / 79.5 * 15.9 = 2.4g of C

Basically, I divided the mass of CuO by the mass of CuO and multiplied by 15.9 to get 15.9g (as in the question). Then, I applied the same equation to C by dividing the mass of C by 79.5 and then multiplying by 15.9 to get 2.4g, which is my answer.

Is this correct?