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  • Stuck here, help me understand: 10 g of hdyrogen and 64 g of oxygen were filled in a steel vessel and exploded. Amount of water produced in this reaction will be:

10 g of hdyrogen and 64 g of oxygen were filled in a steel vessel and exploded. Amount of water produced in this reaction will be:

  • Option 1)

    3 mol

  • Option 2)

    4 mol

  • Option 3)

    1 mol

  • Option 4)

    2 mol

 

Answers (1)

As we learnt

Concept of limiting reagent and excess reagent -

The reactant which gets consumed and thus limits the amount of product formed is called the limiting reagent.

- wherein

 CH4 (g) + 2O2 (g) → CO2 (g) + 2 H2O (g) ;

In this reaction, if 1 mole of methane and 1 mole of oxygen are taken then oxygen would be limiting reagent.

 

 

 The reaction occurs as follows

 

            H_{2}+\frac{1}{2}O_{2} \rightarrow H_{2}O

t=0 \: \: \: \: \: 5 \: \: \: \: \: \: \: \: \: \: 2 \: \: \: \: \: \: \: \: \: 0

t=\infty \: \: \: \: 1 \: \: \: \: \: \: \: \: \: \: 0 \: \: \: \: \: \: \: \: \: 4

No. of moles of hydrogen=10/2=5

No. of moles of oxygen= \frac{64}{32}=2

Taking into account the stoichiometry of the reaction we see,

\frac{1}{2}O_{2} \: requires \rightarrow 1 \: mol \: of \: H_{2}

2O_{2} \: requires \rightarrow 4 \: mol \: of \: H_{2}

Therefore, O2 is the limiting reagent, as 1 mol of H2 will be leftover in the reaction.

Moles of H2O formed from 1/2 mol O2=1

Mole of H2O formed from 2 mol O2= 4 mol

 

 

 


Option 1)

3 mol

This option is incorrect

Option 2)

4 mol

This option is correct

Option 3)

1 mol

This option is incorrect

Option 4)

2 mol

This option is incorrect

Posted by

Vakul

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