Find the volume of 0.240 m sulfuric acid necessary to react completely with 72.8 g sodium hydroxide. express your answer with the appropriate units.

The volume neccesary to complete reaction is 3.79L

Explanation:

First step: Think the equation

H2SO4 + 2NaOH →  Na2SO4  + 2H2O

Afterwards, ballance it.

Relation in reactants is 1:2, so 2 mol of NaOH react to 1 mol of sulfuric.

Gotta find my mol of NaOH

Mass / Molar weight = Mol

72.8 g / 40 g/m = 1.82 mol NaOH

As the relation is 1:2, I have the half of sulfuric: 0.91 mol.

By a rule of three, will be

2 mol NaOH __ react __ 1 mol H2SO4

1.82 mol NaOH __react ___ (1.82 .1)/2

Finally we have to work with molarity, which means mol/L. So, the rule of three is:

0.240 mol H2SO4 __are in __ 1 L

0.91 mol H2SO4___ are in ___ (0.91  .1)/0.240 = 3.79L

i think it is true since you can test the answer and can use observations to complete it.

c. orbital shape i hope this is correct