Science, asked by paramjeetrohilla0001, 5 months ago

. 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

Answered by muktesh57
2

Answer:

0.05mol

this is your answer

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Answered by archanajhaasl
0

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.

\mathrm{n=\frac{m}{M} }                (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;

\mathrm{n=\frac{2.4}{24} =0.1}               (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

#SPJ2

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