Question

1) When 2.38g of magnesium is added to 25.0cm of 2.27 M hydrochloric acid, hydrogen gas is released.

a) Determine the limiting reactant
b) Calculate the mass of hydrogen gas produced.
c) Calculate the mass of excess reactant remained at the end of reaction.
d) What is the percentage yield if 0.044g of hydrogen gas is obtained from the experiment?​

Answer 1

a. HCl.

b. 0.057 g.

c. 1.69 g.

d. 77 %.

Step-by-step explanation:

Hello!

In this case, since the reaction between magnesium and hydrochloric acid is:

`Mg+2HCl\rightarrow MgCl_2+H_2`

Whereas there is 1:2 mole ratio between them.

a) Here, we can identify the limiting reactant as that yielded the fewest moles of hydrogen gas product via the 1:1 and 2:1 mole ratios:

`n_(H_2)^(by\ HCl)=0.025L*2.27(molHCl)/(1L)*(1molH_2)/(2molHCl) =0.0284molH_2\\\\n_(H_2)^(by\ Mg)=2.38gMg*(1molMg)/(24.3gMg)*(1molH_2)/(1molMg)=0.0979molH_2`

Thus, since hydrochloric yields fewer moles of hydrogen than magnesium, we realize it is the limiting reactant.

b) Here, we use the molar mass of gaseous hydrogen (2.02 g/mol) to compute the mass:

`m_(H_2)=0.0284molH_2*(2.02gH_2)/(1molH_2)=0.057gH_2`

c) Here, we compute the mass of magnesium associated with the yielded 0.0248 moles of hydrogen:

`m_(Mg)^(reacted)=0.0284molH_2*(1molMg)/(1molH_2)*(24.3gMg)/(1molMg) =0.690gMg`

Thus, the mass of excess magnesium turns out:

`m_(Mg)^(excess)=2.38g-0.690g=1.69gMg`

d) Finally, we compute the percent yield, considering 0.044 g is the actual yield and 0.057 g the theoretical yield:

`Y=(0.044g)/(0.057g) *100\%\\\\Y=77\%`

Best regards!

Answer 2

a) The limiting reactant would be HCl

From the equation of the reaction:

`Mg (s) + 2 HCl (aq) ---> MgCl_2 (aq) + H_2 (g)`

The mole ratio of Mg to HCl is 1:2.

Mole = mass/molar mass = molarity x volume

Mole of Mg = 2.38/24.3

= 0.098 moles

Mole of HCl = 2.27 x 25/1000

= 0.057 moles

Thus, HCl is limiting while Mg is in excess.

b) Since the mole ratio of HCl to H2 is 2:1:

Mole of H2 produced = 0.057/2

= 0.028 moles

Mass of H2 produced = mole x molar mass

= 0.028 x 2

= 0.057 g

c) Actual mole of Mg that should react = 0.057/2

= 0.028 moles

Excess mole of Mg = 0.098 - 0.028

= 0.07

Mass of excess Mg = 0.07 x 24.3

= 1.701 g

d) Percentage yield if 0.044 g of hydrogen is obtained = yield/theoretical x 100

= 0.044/0.057 x 100

= 77.19%