
4 g of hydrogen is ignited with 4 g of oxygen. The weight of water formed is?
Answer
588.3k+ views
Hint: You can easily find the total amount of ${H_2}O$ generated by just finding out the limiting reagent in the reaction ${H_2} + \dfrac{1}{2}{O_2} \to {H_2}O$. Also, the molecular mass of ${H_2}$= 2 u, molecular mass of ${O_2}$=32 u and molecular mass of ${H_2}O$= 18 u.
Complete step by step solution:
In order to calculate the total amount of ${H_2}O$ formed in the balanced chemical reaction ${H_2} + \dfrac{1}{2}{O_2} \to {H_2}O$ following steps have to be followed:
Step 1: First, we will convert the given amounts into moles.
Mass of ${H_2}$ given = 4 g
Moles of ${H_2}$ $ = \dfrac{{{\text{mass of }}{H_2}}}{{{\text{molecular mass of }}{H_2}}} = \dfrac{4}{2} = 2$
Mass of ${O_2}$ given = 4 g
Moles of ${O_2}$ $ = \dfrac{4}{{32}} = 0.125$.
Step 2: In this step we have to identify limiting reagent.
Limiting reagent is defined as those substances which get completely consumed in a chemical reaction. They also govern the quantity of the final product of a chemical reaction.
From the above balanced equation, we can say that 1 mol of ${H_2}$ reacts with 0.5 mol of ${O_2}$.
Therefore, 2 mol of ${H_2}$ will react with 1 mol of ${O_2}$.
As, we have 0.125 of ${O_2}$, hence ${O_2}$ is the limiting reagent and ${H_2}$ is the excessive reagent
Step 3: In this last step, we will calculate the amount of ${H_2}O$ formed
As the amount of product formed depends on limiting reagent, hence, we can calculate amount of ${H_2}O$ formed as follows : 0.5 mol of ${O_2}$ form ${H_2}O$= 1mol
Therefore, 0.125 mol of ${O_2}$ will form ${H_2}O$ =$\dfrac{{1 \times 0.125}}{{0.5}} = 0.25mol$
Since, molecular mass of ${H_2}O$ is 18 u
Therefore, the mass of ${H_2}O$ formed $ = {\text{ }}0.25 \times 18 = $${\text{4}}{\text{.5 }}g$.
Note: It should be remembered that the chemical reaction should be balanced while finding limiting reagent. Also, the amount of product form in a chemical reaction is limited by limiting reagent.
Also, you should remember that water is formed by mixing hydrogen and oxygen gases and also, a sufficient amount of energy has to be added to provide activation energy to start the reaction.
Complete step by step solution:
In order to calculate the total amount of ${H_2}O$ formed in the balanced chemical reaction ${H_2} + \dfrac{1}{2}{O_2} \to {H_2}O$ following steps have to be followed:
Step 1: First, we will convert the given amounts into moles.
Mass of ${H_2}$ given = 4 g
Moles of ${H_2}$ $ = \dfrac{{{\text{mass of }}{H_2}}}{{{\text{molecular mass of }}{H_2}}} = \dfrac{4}{2} = 2$
Mass of ${O_2}$ given = 4 g
Moles of ${O_2}$ $ = \dfrac{4}{{32}} = 0.125$.
Step 2: In this step we have to identify limiting reagent.
Limiting reagent is defined as those substances which get completely consumed in a chemical reaction. They also govern the quantity of the final product of a chemical reaction.
From the above balanced equation, we can say that 1 mol of ${H_2}$ reacts with 0.5 mol of ${O_2}$.
Therefore, 2 mol of ${H_2}$ will react with 1 mol of ${O_2}$.
As, we have 0.125 of ${O_2}$, hence ${O_2}$ is the limiting reagent and ${H_2}$ is the excessive reagent
Step 3: In this last step, we will calculate the amount of ${H_2}O$ formed
As the amount of product formed depends on limiting reagent, hence, we can calculate amount of ${H_2}O$ formed as follows : 0.5 mol of ${O_2}$ form ${H_2}O$= 1mol
Therefore, 0.125 mol of ${O_2}$ will form ${H_2}O$ =$\dfrac{{1 \times 0.125}}{{0.5}} = 0.25mol$
Since, molecular mass of ${H_2}O$ is 18 u
Therefore, the mass of ${H_2}O$ formed $ = {\text{ }}0.25 \times 18 = $${\text{4}}{\text{.5 }}g$.
Note: It should be remembered that the chemical reaction should be balanced while finding limiting reagent. Also, the amount of product form in a chemical reaction is limited by limiting reagent.
Also, you should remember that water is formed by mixing hydrogen and oxygen gases and also, a sufficient amount of energy has to be added to provide activation energy to start the reaction.
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