Question #957e1

1 Answer
Feb 29, 2016

Answer:

A) #6.0*10^22# molecules of #CO_2#
B) #6.0*10^22# molecules #O_2#

Explanation:

You are given 4.4g of #CO_2# gas. To calculate the molecules of #CO_2# present, you need to first covert from grams of #CO_2# to moles of #CO_2#.

4.4g #CO_2# *#(1 mol)/(44.1g)#= .10 mol of #CO_2#

After you covert to moles of #CO_2#, you then covert to molecules of #CO_2# with Avogadro's number; #6.022 * 10^23#,

.1 mol #CO_2# #(6.022*10^23)/(1 mol)#= #6.0*10^22# molecules of #CO_2#

to do all your conversion in one step;

4.4g #CO_2# #(1mol)/(44.1g)# #(6.022*10^23)/(1mol)#= #6.0*10^22# molecules of #CO_2#

To determine the atoms of oxygen present;

You are given 4.4g #CO_2#, but this time we want atoms (or molecules) of Oxygen gas.

4.4g #CO_2#* #(1mol)/(44.1g)#= .10 mol #CO_2#

For every mol of #CO_2#, there is 1 mol of oxygen #(O_2)#

.10 mol #CO_2# #(1mol O_2)/(1mol CO_2)#=.10 mol #O_2#

we use Avogadro's number again;

.10 mol #O_2# #(6.022*10^23)/(1mol O_2)#= #6.0*10^22# molecules/atoms #O_2#

All worked out;

4.4g #CO_2# #(1mol CO_2)/(44.1g CO_2)# #(1mol O_2)/(1mol CO_2)# #(6.022*10^23)/(1mol O_2)#= #6.0*10^22# molecules of #O_2#