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

Answer:

The correct answer is option 3.

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:

$1\times 1.55 M\times V_1=2\times 4.66 M\times 45.8 mL$

$V_1=\frac{2\times 4.66 M\times 45.8 mL}{1\times 1.55 M}=275 mL$

1 mL = 0.001 L

$V_1=275\times 0.001 L=0.0.275 L$

Hence, the correct answer is option 3.