Maximum number of moles of potash alum $K_{S} O_{4} . A l_{2} (S O_{4})_{3} .24 H_{2} O$ can be made from sample of aluminium containing $5.4$ g of aluminium (considering $100$ yield ) is:

10^{- 2}

mole

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0.5 \times 10^{- 2}

mole

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0.2

moles

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0.1

mole

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Solution

The correct option is D $0.1$ mole
It is clear that $2$ moles of $A l$ will be required to produce $1$ mole of potash alum.
Given moles of $A l = \frac{5.4}{27} = 0.2$
$∴$ Moles of potash alum produced $= 0.1$

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Similar questions

Q. Maximum number of moles of potash alum

K_{2} S O_{4} A l_{2} (S O_{4})_{3}

can be made from sample of aluminum-containing

5.4 g

of aluminium (considerning 100% yield) is?

25.4 g of iodine and 14.2 g of chlorine are made to react completely to yield a mixture of ICl and $I C l_{3}$ . Calculate the number of moles of ICl and $I C l_{3}$ is formed

Q. How many moles of ferric alum

(N H_{4})_{2} S O_{4} . F e_{2} (S O_{4})_{3} .24 H_{2} O

can be made from the sample of

F e

containing

0.0056

g of it?

Q. How many moles of ferric alum (NH4)2SO4Fe2(SO4)3 24H2O can be made from the sample of Fe containing 0.0056g of it?

Q. The total number of atoms present in

1

mole of potash alum, in terms of Avogadro number

N_{A}

, is:

[ Potash Alum

= K_{2} S O_{4} . A l_{2} (S O_{4})_{3} . 24 H_{2} O

]

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Maximum number of moles of potash alum KSO4.Al2(SO4)3.24H2O can be made from sample of aluminium containing 5.4g of aluminium (considering 100 yield ) is:

The correct option is D 0.1 moleIt is clear that 2 moles of Al will be required to produce 1 mole of potash alum.Given moles of Al=5.427=0.2∴ Moles of potash alum produced =0.1

Maximum number of moles of potash alum $K_{S} O_{4} . A l_{2} (S O_{4})_{3} .24 H_{2} O$ can be made from sample of aluminium containing $5.4$ g of aluminium (considering $100$ yield ) is:

The correct option is D $0.1$ mole
It is clear that $2$ moles of $A l$ will be required to produce $1$ mole of potash alum.
Given moles of $A l = \frac{5.4}{27} = 0.2$
$∴$ Moles of potash alum produced $= 0.1$