Question #ef574

At first you need to have the whole balanced equation:
#H_2SO_4 + 2NaOH -> Na_2SO_4 + 2H_2O#
It is a neutralisation reaction and those have regular mechanism.

First we need to know how many moles of #H_2SO_4# we need.
For this we can use cross-multiplication
From the reaction we've written before we can see that 2 moles of #NaOH# react with one mole of #H_2SO_4# (no number means 1 mole)

# 1 H_2SO_4 + 2NaOH ->1 Na_2SO_4 + 2H_2O#

So we use the cross multiplication
2 moles #NaOH # ..................................1 mole #H_2SO_4 #
0.850 moles #NaOH#............................x mole #H_2SO_4#

#2*x= 1*0.850//:2 #
#x=0.425# moles

Now that we have number of moles, we just need the molar mass (equation: #n=m/M # , where n= number of moles, m = mass, M = molar mass)

#M=2+(4*16)+32=98 g//mol #
After rearranging the equation: #m=n*M#
#m=0.425*98#
#m=41.6g#

If anything is unclear, feel free to ask :)

First write down the reaction in full:

#H_2SO_4+2NaOHrarr Na_2SO_4+2H_2O#.

We won't need the equation, but it is clearer if you do so.

Notice how the moles of sulfuric acid and sodium hydroxide are in the ratio #2:1#. So if we have #0.85"mol"# of #NaOH#:

#(2 "mol "NaOH)/(1 "mol " H_2SO_4)=(0.85"mol " NaOH)/(x)#

Cross-multiplying:

#1 "mol " H_2SO_4*0.85"mol " NaOH=x*2 "mol "NaOH#

#x=(1 "mol " H_2SO_4*0.85cancel("mol " NaOH))/(2 cancel("mol "NaOH))#

#x=0.425"mol " H_2SO_4#

The formula for moles is #n=m/M#, where #n# is the moles, #m# the mass, and #M# the molar mass.

Rearranging to solve for #m#, we get:

#m=nM#.

Since we have #0.425"mol"# of sulfuric acid, and the molar mass of the acid is #98.08"g"/"mol"#.

We can input our values into the rearranged equation:

#m=0.425cancel("mol")*98.08"g"/cancel("mol")#

#m=41.68"g"# of sulfuric acid reacts with #0.85# moles of sodium hydroxide to produce water and sodium sulfate.