Question

Calculate the volume of H2(g) at 273 K and 2.00 atm that will be formed when 275 mL of 0.725 M HCl solution reacts with excess Mg to give hydrogen gas and aqueous magnesium chloride

Answer 1

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

Answer 2

Answer:

$V=1.11L$

Explanation:

Hello,

In this case, the chemical reaction is:

$2HCl+Mg\rightarrow MgCl_2+H_2$

Thus, the reacting moles of hydrochloric acid are computed with the volume of the given solution:

$n_{HCl}=0.725\frac{mol}{L} *0.275L=0.199molHCl$

Hence, we perform the proportional factors having the 2:1 molar relationship between hydrochloric acid and hydrogen to find the yielded moles of hydrogen:

$n_{H_2}=0.199molHCl*\frac{1molH_2}{2molHCl} =0.0995molH_2$

Finally, by using the ideal gas equation we obtain the volume for the specified conditions as shown below:

$V=\frac{nRT}{P} =\frac{0.0995molH_2*0.082\frac{atm*L}{mol*K}*273K}{2.00atm} \\\\V=1.11L$

Best regards.