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

1 Answer
May 9, 2016

Answer:

#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#