What mass of sodium chloride, #NaCl#, is made when #200# #L# of chlorine gas, #Cl_2#, reacts with excess sodium metal?

1 Answer
May 21, 2017

Answer:

#1044# #g# of #NaCl# are produced, since there are #8.93# mol of #Cl_2# gas and each mole of #Cl_2# produces #2# mol of #NaCl#.

Explanation:

One mole of an ideal gas has a volume of #22.4# #L# at STP. Chlorine is not a perfectly ideal gas but can be treated as one at this temperature and pressure.

The number of moles of gas, then, is given by #n=V/22.4 = 200/22.4 = 8.93# #mol#

Looking at the balanced equation, each #1# mol of #Cl_2# yields #2# mol of #NaCl#, so #8.93# mol of #Cl_2# yields #2xx8.93 =17.86# mol of #NaCl#.

The molar mass of #NaCl# is #58.44# #g#. To find the mass of the product, we take #n=m/M# and rearrange to make #m# the product. #m=nM = 17.86xx58.44 = 1044# #g# (with some rounding)