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?
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:
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:
\(n_1=1\\M_1=1.55 M\\V_1=\ _2=2\\M_2=4.66 M\\V_2=45.8 mL\)
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.