Answer:
a. HCl.
b. 0.057 g.
c. 1.69 g.
d. 77 %.
Explanation:
Hello!
In this case, since the reaction between magnesium and hydrochloric acid is:
[tex]Mg+2HCl\rightarrow MgCl_2+H_2[/tex]
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:
[tex]n_{H_2}^{by\ HCl}=0.025L*2.27\frac{molHCl}{1L}*\frac{1molH_2}{2molHCl} =0.0284molH_2\\\\n_{H_2}^{by\ Mg}=2.38gMg*\frac{1molMg}{24.3gMg}*\frac{1molH_2}{1molMg}=0.0979molH_2[/tex]
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:
[tex]m_{H_2}=0.0284molH_2*\frac{2.02gH_2}{1molH_2}=0.057gH_2[/tex]
c) Here, we compute the mass of magnesium associated with the yielded 0.0248 moles of hydrogen:
[tex]m_{Mg}^{reacted}=0.0284molH_2*\frac{1molMg}{1molH_2}*\frac{24.3gMg}{1molMg} =0.690gMg[/tex]
Thus, the mass of excess magnesium turns out:
[tex]m_{Mg}^{excess}=2.38g-0.690g=1.69gMg[/tex]
d) Finally, we compute the percent yield, considering 0.044 g is the actual yield and 0.057 g the theoretical yield:
[tex]Y=\frac{0.044g}{0.057g} *100\%\\\\Y=77\%[/tex]
Best regards!
a) The limiting reactant would be HCl
From the equation of the reaction:
[tex]Mg (s) + 2 HCl (aq) ---> MgCl_2 (aq) + H_2 (g)[/tex]
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%
