Respuesta :
Answer:
For 1: The volume of HCl required is 6 L.
For 2: The volume of HCl required is 9 L.
For 3: The volume of sulfuric acid required is 4.5 L.
Explanation:
To calculate the volume of acid, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex] ......(1)
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base
- For 1:
We are given:
[tex]n_1=1\\M_1=1M\\V_1=?L\\n_2=1\\M_2=3.0M\\V_2=2.0L[/tex]
Putting values in equation 1, we get:
[tex]1\times 1\times V_1=1\times 3\times 2\\\\V_1=6L[/tex]
Hence, the volume of HCl required is 6 L.
- For 2:
We are given:
[tex]n_1=1\\M_1=1M\\V_1=?L\\n_2=2\\M_2=3.0M\\V_2=1.5L[/tex]
Putting values in equation 1, we get:
[tex]1\times 1\times V_1=2\times 3.0\times 1.5\\\\V_1=9L[/tex]
Hence, the volume of HCl required is 9 L.
- For 3:
To calculate the volume of acid, we use the equation:
[tex]N_1V_1=N_2V_2[/tex]
where,
[tex]N_1\text{ and }V_1[/tex] are the normality and volume of acid
[tex]N_2\text{ and }V_2[/tex] are the normality and volume of base
We are given:
[tex]N_1=1.0N\\V_1=?L\\N_2=3.0N\\V_2=1.5L[/tex]
Putting values in above equation, we get:
[tex]1.0\times V_1=3.0\times 1.5\\\\V_1=4.5L[/tex]
Hence, the volume of sulfuric acid required is 4.5 L.
