Question

42.
(4) 4
(3) 3 4
34. 2.4 g of magnesium is burnt with excess of O,
gas, the amount of magnesium oxide formed will
1
be (Atomic mass of Mg = 24 u)
(2) 0.01 mole
(1) 0.1 mole
(4) 0.025 mole
(3) 0.05 mole
35. A metal oxide containing 70% of metal (At. wt of
metal = 56) and rest oxygen. Its empirical formula
would be
(1) MO
(2) MO,
(3) MO,
(4) MO
36. Mass percentage of sulphur in So, is (Atomic​

Answer 1

Answer :

(34) The moles of magnesium oxide will be, 0.1 mole and amount of magnesium oxide will be, 4 grams.

(35) The empirical formula will be, $M_2O_3$

Solution for 34 question :

First we have to calculate the moles of Mg.

$\text{Moles of }Mg=\frac{\text{Mass of }Mg}{\text{Molar mass of }Mg}=\frac{2.4g}{24g/mole}=0.1moles$

Now we have to calculate the moles of MgO.

The balanced chemical reaction is,

$2Mg+O_2\rightarrow 2MgO$

From the balanced reaction we conclude that

As, 2 moles of $Mg$ react to give 2 moles of $MgO$

So, 0.1 moles of $Mg$ react to give 0.1 moles of $MgO$

The moles of MgO = 0.1 mole

Now we have to calculate the mass of MgO.

$\text{Mass of }MgO=\text{Moles of }MgO\times \text{Molar mass of }MgO$

$\text{Mass of }MgO=(0.1mole)\times (40g/mole)=4g$

The amount of MgO = 4 grams

Solution for 35 question :

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of M = 70 % = 70 g

Mass of O = 100 - 70 = 30 g

Molar mass of M = 56 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of M = $\frac{\text{ given mass of M}}{\text{ molar mass of M}}= \frac{70g}{56g/mole}=1.25moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{30g}{16g/mole}=1.875moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For M = $\frac{1.25}{1.25}=1$

For O = $\frac{1.875}{1.25}=1.5$

Now multiply the number of M and O by 2, we get the whole number ratio.

The ratio of M : O = 2 : 3

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $M_2O_3$

Therefore, the empirical formula will be, $M_2O_3$