How many particle or molecules of #Na_2CO_3# are produced when 10g of #NaOH# is used? #22.3 x 10^24# particles of #NaOH# are needed to produce how many liters of #H_2O#? How many particles/molecules of #NaOH# are from 30 gram of #NaOH#?

#2NaOH(s) + CO_2(g) -> Na_2CO_3(s) + H_2O(l)#

1 Answer
Aug 2, 2017

Answer:

a) #7.5 xx 10^22 # molecules of #Na_2CO_3# are produced
b)cannot be answered as stoichiometry deals with gases not liquids and water is liquid
c)There are #4.5xx 10^23 # molecules of #NaOH# in #30 g#

Explanation:

#2NaOH(s) + CO_2(g) -> Na_2CO_3(s) + H_2O(l)#

Molar mass of #NaOH = 40 #

#2xx40 g# of #NaOH# produces #6.02 xx 10^23# molecules of #Na_2CO_3#
#:.# #10 g# of #NaOH# produces #(6.02xx10^23xx10)/(2xx40)# molecules of #Na_2CO_3#
#= 7.5 xx 10^22# molecules of #Na_2CO_3#

#40g# of #NaOH# contains #6xx10^23# molecules
#:.# #30g# of #NaOH# contains #(30xx6xx10^23)/40# molecules
#=4.5xx10^23# molecules