Question

3.42g of sucrose are dissolved in 18 g of water in a beaker . the no of oxygen atoms in the solution are

Answer 1

C12H22O11 + H2O --> 4C2H5OH + 4CO2
(sucrose)
No of moles of C12H22O11=GIVEN MASS/MOLAR MASS
                                            =3.42/342 = 0.01
According to the reaction,
1 mol of C12H22O11 requires 1 mol of H2O
But we have only 0.01 moles of C12H22O11. Hence, C12H22O11 is a limiting reagent.
So, C2H5OH will be formed only from amount of available sucrose.
From the reaction,
0.01 mol of C12H22O11 forms 4 moles of O   (solution formed-C2H50H)
==> 1mol of C12H22O11 forms=0.01*4=0.04mol of O
Now,
Since 1 mol of O contains 6.022*10^23 atoms
So, 0.04 mol of O will have=6.022*10^23 * 0.04=2.4 * 10^22 atoms of O.

Answer 2

C12H22O11 + H2O --> 4C2H5OH + 4CO2

(sucrose)

No of moles of C12H22O11=GIVEN MASS/MOLAR MASS

                                            =3.42/342 = 0.01

According to the reaction,

1 mol of C12H22O11 requires 1 mol of H2O

But we have only 0.01 moles of C12H22O11. Hence, C12H22O11 is a limiting reagent.

So, C2H5OH will be formed only from amount of available sucrose.

From the reaction,

0.01 mol of C12H22O11 forms 4 moles of O   (solution formed-C2H50H)

==> 1mol of C12H22O11 forms=0.01*4=0.04mol of O

Now,

Since 1 mol of O contains 6.022*10^23 atoms

So, 0.04 mol of O will have=6.022*10^23 * 0.04=2.4 * 10^22 atoms of O.