Question

What is the volume of 36.8 g of carbon monoxide at STP?

Answer 1

`V=29.4L`

Explanation:

Let us first find the number of mole of carbon monoxide in `36.8g`:

`n=m/(MM)=(36.8cancel(g))/(28.0cancel(g)/(mol))=1.31mol`

At STP, the molar volume is equal to `22.4L/(mol)`, therefore, the volume that will be occupied by `36.8g` carbon monoxide is:

`V=1.31cancel(mol)xx(22.4L)/(1cancel(mol))=29.4L`