Question

For the neutralization reaction involving HNO3 and Ca(OH)2, how many liters of 1.55 M HNO3 are needed to react with 45.8 mL of a 4.66 M Ca(OH)2 solution?

1. 0.137 L 2. 0.0343 L 3. 0.275 L 4. 1.32 L 5. 0.662 L 6. 0.330 L

Answer 1

2HNO3 + Ca(OH)2 ==> Ca(NO3)2 + 2H2O
mols Ca(OH)2 = M x L = ?
Using the coefficients in the balanced equation, convert mols Ca(OH)2 to mols HNO3.
Then M HNO3 = mols HNO3/LHNO3. You have mols and M, solve for L.

Answer 2

The correct answer is option 3.

Step-by-step explanation:

`2HNO_3(aq)+Ca(OH)_2\rightarrow Ca(NO_3)_2(aq)+2H_2O(l)`

To calculate the concentration of acid, we use the equation given by neutralization reaction:

`n_1M_1V_1=n_2M_2V_2`

where,

`n_1,M_1\text{ and }V_1` are the n-factor, molarity and volume of acid which is
`HNO_3`

`n_2,M_2\text{ and }V_2` are the n-factor, molarity and volume of base which is
`Ca(OH)_2`.

We are given:

`n_1=1\\M_1=1.55 M\\V_1=\\n_2=2\\M_2=4.66 M\\V_2=45.8 mL`

Putting values in above equation, we get:

`1* 1.55 M* V_1=2* 4.66 M* 45.8 mL`

`V_1=(2* 4.66 M* 45.8 mL)/(1* 1.55 M)=275 mL`

1 mL = 0.001 L

`V_1=275* 0.001 L=0.0.275 L`

Hence, the correct answer is option 3.