CameraIcon

SearchIcon

MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Limiting Reagent or Reactant

Which of the ...

Question

Which of the following statements is/are wrong for the following reactions?

$P_{4} + 5 O_{2} \to P_{4} O_{10}$
$P_{4} + 3 O_{2} \to P_{4} O_{6}$

$1.24$ g of $P_{4}$ reacts with $8.0$ g of $O_{2}$ .

A

P_{4}

is the limiting quantity

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B

O_{2}

is the limiting quantity

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

C

Mass of

P_{4} O_{10}

obtained is

2.2

g

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

D

Mass of

P_{4} O_{6}

obtained is

2.84

g

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct options are
B $O_{2}$ is the limiting quantity
C Mass of $P_{4} O_{10}$ obtained is $2.2$ g
D Mass of $P_{4} O_{6}$ obtained is $2.84$ g
$n_{P_{4}} = \frac{1.24}{31 \times 4} = 0.01 m o l, n_{O_{2}} = \frac{8}{32} = 0.25 m o l$

In reaction (i),
Moles of $O_{2}$ required $= (0.01 m o l P_{4}) (\frac{5 m o l O_{2}}{m o l P_{4}})$

$= 0.01 \times 5 = 0.05 m o l$

Since $O_{2}$ is in excess amount.

Therefore, $P_{4}$ is the limiting quantity.

Molecular weight $P_{4} O_{10} = 284 g$

0.01 mol of $P_{4}$ produces $=$ 0.01 mol of $P_{4} O_{10}$

So $P_{4} O_{10}$ in terms of mass $= 0.01 \times 284 = 2.84 g$

Molecular weight of $P_{4} O_{6} = 220$

Weight of $P_{4} O_{6}$ produced $= (0.01 m o l P_{4}) (\frac{1 m o l P_{4} O_{6}}{m o l P_{4} O_{6}}) \times M_{P_{4} O_{6}}$
$= 0.01 \times 220 = 2.2 g$

Hence options B,C & D are correct.

Suggest Corrections

thumbs-up

2

Similar questions

Q. Two substance,

P_{4}

O_{2}

are allowed to react completely to form mixture of

P_{4} O_{6}

P_{4} O_{10}

leaving none of the reactants. The composition of final mixture if

P_{4}

O_{2}

P_{4} + 3 O_{2} \to P_{4} O_{6}

P_{4} + 5 O_{2} \to P_{4} O_{10}

.

1

P_{4}

4

O_{2}

3

P_{4}

11

O_{2}

3

P_{4}

13

O_{2}

Q. What weights of

P_{4} O_{6}

P_{4} O_{10}

will be produced by the combustion of 31 g of

P_{4}

with 32g of oxygen leaving no

P_{4}

O_{2}

?

Q. A reaction container holds 5.77 g of

P_{4}

O_{2}

and the following reaction occurs:

P_{4} + O_{2} \to P_{4} O_{6} .

The amount of excess reagent left is?

Q. What weights of

P_{4} O_{6}

P_{4} O_{10}

will be produced by the combustion of

2 g

P_{4}

2 g

of oxygen leaving no

P_{4}

O_{2}

P_{4} O_{10}

440

P_{4} S_{3}

384

O_{2}

P_{4} S_{3} + O_{2} ⟶ P_{4} O_{10} + {S O}_{2}

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Limiting Reagent

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Limiting Reagent or Reactant

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon