Question #2b66a

1 Answer
May 7, 2016

Answer:

#2NaOH(aq) + H_2SO_4(aq) rarr Na_2SO_4(aq) + 2H_2O(l)#

Explanation:

The given equation is absolutely vital to answering this question. It says that 2 mol sodium hydroxide, approx. #80.0*g#, react with the 1 mol sulfuric acid, a mass of approx. #98.08*g#, to give stoichiometric quantities of sodium sulfate and water.

So we must calculate the molar quantities of our starting materials.

#"Moles of sodium hydroxide"=(20.0*g)/(40.00*g*mol^-1)# #=# #0.5*mol#.

#"Moles of sulfuric acid"=(9.8*g)/(98.08*g*mol^-1)# #=# #0.1*mol#.

So, clearly, sulfuric acid is the limiting reagent. And at most, I can get #0.2*mol# of #NaOH# to react.

Given this, only #0.1*mol# of #Na_2SO_4# will be formed, i.e. #0.1*molxx142.04*g*mol^-1# #=# #14.2*g#.

All of these calculations relies on the given stoichiometric reaction. If I don't have this, or if I have not balanced it correctly, I will not be able to do the calculation. Capisce?