How many grams of NaF form when 25.7g of HF reacts with excess sodium silicate?

1 Answer
Jan 12, 2016

Answer:

#2HF(aq) + Na_2SiO_3(aq) rarr 2NaFdarr + H_2SiO_3(aq)#

Explanation:

The stoichiometry of the equation is crucial; inasmuch as the stoichiometry, the chemical proportion, shows that hydrogen fluoride and sodium fluoride are formed in equal amounts, and all I need are the molecular weights of hydrogen fluoride (#20.0*g*mol^-1)# and sodium fluoride (#42.0*g*mol^-1)#.

So moles of #HF# #=# #(25.7*g)/(20.0*g*mol^-1)# #=# #??# #mol#

Moles of #NaF# #=# #(20.0*g)/(25.7*g*mol^-1) xx42.0*g*mol^-1# #=# #??# #g#