Respuesta :
v=4.8 L
c=5.0 mol/L
M(Mg)=24.3 g/mol
1) n(HCl)=cv
2) m(Mg)=M(Mg)n(HCl)/2
3) m(Mg)=M(Mg)cv/2
m(Mg)=24.3*5*4.8/2=291.6 g
c=5.0 mol/L
M(Mg)=24.3 g/mol
1) n(HCl)=cv
2) m(Mg)=M(Mg)n(HCl)/2
3) m(Mg)=M(Mg)cv/2
m(Mg)=24.3*5*4.8/2=291.6 g
Answer:
291.72 grams of magnesium metal will react completely with 4.8 liters of 5.0 M HCl.
Explanation:
[tex]Mg (s) + 2HCl (aq) \rightarrow MgCl_2(aq)+H_2(g)[/tex]
Concentration of 4,8 L HCl solution = 5.0 M
Moles of HCl in solution = n
[tex]Molarity=\frac{Moles}{Volume (L)}[/tex]
[tex]5.0 M=\frac{n}{4.8L}[/tex]
n = 24 moles
According to reaction , 2 moles of HCl reacts with 1 mole of magnesium metal.
Then 24 moles of HCl will react with :
[tex]\frac{1}{2}\times 24 = 12 mol[/tex] of magnesium
Mass of 12 moles of magnesium = 12 mol × 24.31 g/mol=291.72g
291.72 grams of magnesium metal will react completely with 4.8 liters of 5.0 M HCl.
