Answer:
[tex]V=1.11L[/tex]
Explanation:
Hello,
In this case, the chemical reaction is:
[tex]2HCl+Mg\rightarrow MgCl_2+H_2[/tex]
Thus, the reacting moles of hydrochloric acid are computed with the volume of the given solution:
[tex]n_{HCl}=0.725\frac{mol}{L} *0.275L=0.199molHCl[/tex]
Hence, we perform the proportional factors having the 2:1 molar relationship between hydrochloric acid and hydrogen to find the yielded moles of hydrogen:
[tex]n_{H_2}=0.199molHCl*\frac{1molH_2}{2molHCl} =0.0995molH_2[/tex]
Finally, by using the ideal gas equation we obtain the volume for the specified conditions as shown below:
[tex]V=\frac{nRT}{P} =\frac{0.0995molH_2*0.082\frac{atm*L}{mol*K}*273K}{2.00atm} \\\\V=1.11L[/tex]
Best regards.
Answer:
The volume of H2 is 1.117 L
Explanation:
Step 1: Data given
Temperature of H2 = 273 K
Pressure of H2 = 2.00 atm
Molarity of HCl = 0.725 M HCl
Volume of HCl = 275 mL = 0.275 L
Mg is in excess
Step 2: The balanced eqquation
Mg + 2HCl → MgCl2 + H2
Step 3: Calculate moles HCl
Moles HCl = molarity * volume
Moles HCl = 0.725 M * 0.275 L
Moles HCl = 0.199375 moles
Step 4: Calculate moles H2
For 1 mol Mg we need 2 mole HCl to produce 1 mol MgCl2 and 1 mol H2
For 0.199375 moles HCl we'll have 0.199375/2 = 0.0996875 moles H2
Step 5: Calculate volume H2
p*V = n*R*T
⇒with p = the pressure of H2 = 2.0 atm
⇒with V = the volume of H2 = TO BE DETERMINED
⇒with n = the moles of H2 = 0.0996875 moles
⇒with R = the gas constant = 0.08206 L*atm/mol*K
⇒with T = the temperature = 273 K
V = (n*R*T) / p
V = (0.0996875 * 0.08206 * 273) / 2.0
V = 1.117 L
The volume of H2 is 1.117 L
