Question

A mixture contains 5.4 g of AI, 1.2 g of Mg and
4.6 g of C2H5OH. The ratio of their moles is
(at. wt. of Al=27 and Mg = 24)

(1) 4:1:2
(2) 2:1:5
(3) 2:1:4
(4) 2 :3 :4​

Answer 1

Answer:-

Given:

Mass of Aluminium (Al) = 5.4 g

Molar mass of Al = 27 g/mol

Mass of Magnesium (Mg) = 1.2 g

Molar mass of Mg = 24 g/mol

Mass of Ethanol (C2H5OH) = 4.6 g

Molar mass of C2H5OH = 2(12) + 5(1) + (16) + 1 = 45 g/mol.

We know that,

n ( Number of moles ) = Mass in gms/Molar mass.

No. of moles of Al = 5.4/27 = 0.2

No. of moles of Mg = 1.2/24 = 0.05

No. of moles of Ethanol = 4.6/45 = 0.1

Ratio of number of moles of Al to Mg = 0.2/0.05 = 4

Ratio of number of moles of Mg to C2H5OH = 0.05/0.1 = 0.5 = 1.

Ratio of number of moles of Al to C2H5OH = 0.2/0.1 = 2

Hence, the ratio of number of moles of Al , Mg , C2H5OH is 4 : 1 : 2.