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

1 Answer
May 9, 2016

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