# 0.850 L of 0.490 M #H_2SO_4# is mixed with 0.800 L of 0.280 M #KOH#. What concentration of #H_2SO_4# remains after neutralization?

##### 1 Answer

The concentration of

#### Explanation:

This is really a two-part question:

- How much of the
#"H"_2"SO"_4# is neutralized? - What is the concentration of the remaining
#"H"_2"SO"_4# ?

**1. How much of the #"H"_2"SO"_4# is neutralized?**

This is a volume-moles stoichiometry problem.

The steps are:

a. Write the balanced equation.

b. Use the molarity of

c. Use the molar ratio from the balanced equation to convert moles of

d. Use the molarity to convert the initial volume of

e. Calculate the moles of excess

**a. Write the balanced equation:**

**b. Calculate the moles of #"KOH"#.**

**c. Calculate the moles of #"H"_2"SO"_4# that reacted**

The molar ratio of

∴

**2. What is the concentration of the remaining #"H"_2"SO"_4#?**

**a. Calculate the initial moles of #"H"_2"SO"_4#.**

**e. Calculate the moles of #"H"_2"SO"_4# remaining**

**e. Calculate the molarity of the excess #"H"_2"SO"_4#**

∴