. 2.4 g of pure Mg (at. mass 24) is dropped in
100 mL of 1M HCl. Calculate mol of Mg is left behind
Answers
Answer:
0.05mol
this is your answer
Answer:
The mole of Mg that is left behind is 0.05 mol.
Explanation:
First, let's see how the reaction will proceed.
Mg+2HCl⇒MgCl₂+H₂ (1)
Where,
Mg=magnesium
HCl=hydrogen chloride
MgCl₂=magnesium chloride
H₂=hydrogen
Now let's find the number of moles of magnesium.
(2)
n=number of moles
m=mass of the given element
M=molar mass of the element
From the question we have,
The given mass of magnesium is(m)=2.4 g
The molar mass of magnesium(M)=24 g
Volume of the solution(V)=100mL
The molarity of HCl=1M
By inserting the values in equation (2) we get;
(3)
100mL of 1M HCl=0.1 mol HCl (4)
From equation (1) we can see that 1 mol of Mg reacts with 0.2 mol of HCl.
So, HCl is the limiting reagent here.
Thus the excess amount or left behind amount of Mg =0.05 mol
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