Question

1) What is the mass of oxygen required to react completely with $ 24{\text{ }}g $ of $ C{H_4} $ in the following reaction?
 $ C{H_4}\left( g \right) + 2{O_2}\left( g \right){\text{ }} \to C{O_2}\left( g \right) + 2{H_2}O\left( g \right) $
2) How much mass of $ C{H_4} $ would react with $ 96{\text{ }}g $ of oxygen?

Answer 1

Hint :The question are provided with the number of moles and mass is also given simply calculate the mass first with the formula $ \dfrac{{given{\text{ }}mass}}{{molar{\text{ }}mass}} $ and solve the problem then in next question use the same formula but carefully first calculate the number of moles.

Complete Step By Step Answer:
1. In the first question we are given that $ 24{\text{ }}g $ of $ C{H_4} $ is reacting with a certain mass of oxygen in the reaction. We have to find how much oxygen is reacting. The reaction taking place here is:
 $ C{H_4}\left( g \right) + 2{O_2}\left( g \right){\text{ }} \to C{O_2}\left( g \right) + 2{H_2}O\left( g \right) $
If we look into the reaction carefully we can see that $ 1 $ mole of $ C{H_4} $ reacts with $ 2 $ moles of oxygen and releases $ 1 $ mole of $ C{O_2} $ and $ 2 $ moles of $ {H_2}O $ . This means that $ 1 $ mole of $ C{H_4} $ will require $ 2 $ moles of oxygen for the reaction to take place.
It is given in the question that $ 24{\text{ }}g $ of $ C{H_4} $ is reacting. It is known that $ 1 $ mole of $ C{H_4} = 12 + 4 \times 1 = 16g $ , so moles of $ C{H_4} $ present in the reaction is $ \dfrac{{given{\text{ }}mass}}{{molar{\text{ }}mass}} $
Given mass of $ C{H_4} $ is $ 24{\text{ }}g $ and molar mass is $ 16{\text{ }}g $
Hence
 $ \Rightarrow $ $ \dfrac{{given{\text{ }}mass}}{{molar{\text{ }}mass}} = \dfrac{{24}}{{16}} = 1.5moles $
From the equation $ 1 $ mole of $ C{H_4} $ requires $ 2 $ moles of oxygen thus $ 1.5 $ moles of $ C{H_4} $ will require $ 2 \times 1.5 = 3 $ moles of oxygen.
Now it is also known that $ 1 $ mole of oxygen $ = 16 $ g thus $ 3 $ moles of oxygen will be equal to $ 3 \times 16 = 48 $ g.
Therefore the mass of oxygen required to react completely with $ 24{\text{ }}g $ of $ C{H_4} $ is $ 48 $ g.
2. In the second question it is said that $ 96 $ g of oxygen is reacting with a certain g of $ C{H_4} $ we need to find the mass of $ C{H_4} $ which is reacting.
The equation we have with us is : $ C{H_4}\left( g \right) + 2{O_2}\left( g \right){\text{ }} \to C{O_2}\left( g \right) + 2{H_2}O\left( g \right) $
In the above reaction $ 1 $ mole of $ C{H_4} $ reacts with $ 2 $ moles of oxygen and releases $ 1 $ mole of $ C{O_2} $ and $ 2 $ moles of $ {H_2}O $ . This means that $ 2 $ moles of oxygen will require $ 1 $ mole of $ C{H_4} $ for the reaction to take place.
Given mass of oxygen is $ 96 $ g and molar mass of oxygen is $ 32 $ g.
So $ \dfrac{{given{\text{ }}mass}}{{molar{\text{ }}mass}} = \dfrac{{96}}{{32}} = 3 $ moles.
We see that $ 2 $ moles of oxygen will require $ 1 $ mole of $ C{H_4} $ . So $ 1 $ mole of oxygen will require $ \dfrac{1}{2} $ moles of $ C{H_4} $ and so $ 3 $ moles of oxygen will require $ \dfrac{3}{2} $ moles of $ C{H_4} $ .
thus moles of $ C{H_4} = \dfrac{3}{2} $
we know that No. of moles = $ \dfrac{{given{\text{ }}mass}}{{molar{\text{ }}mass}} $
so, No. of moles of $ C{H_4} $ $ = \dfrac{{given{\text{ }}mass}}{{16}} $
 Mass of $ C{H_4}{\text{ }} = {\text{ }}\dfrac{3}{2} \times 16 = 24{\text{ }}g $
Therefore $ 96{\text{ }}g $ of oxygen would require $ 24 $ g of $ C{H_4} $ .

Note :
These questions are very easy to solve only you have to see how the steps are being followed in the above solutions. Always remember to take into account the number of moles of molecules that are taking part in the raction.