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?

10^{- 2}

mole

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0.5 \times 10^{- 2}

moles

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0.2

moles

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0.1

mole

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Solution

The correct option is B $0.1$ mole
We have,
molar mass of potash alum = 474.3 g/mol
mass of aluminium present in 1 mole of potash alum = 54 g
so, moles of potash alum prepared by 5.4 g of aluminium = $\frac{1}{54} * 5.4 = 0.1 m o l e$ .

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Maximum number of moles of potash alum K2SO4Al2(SO4)3 can be made from sample of aluminum-containing 5.4g of aluminium (considerning 100% yield) is?

The correct option is B 0.1 moleWe have,molar mass of potash alum = 474.3 g/molmass of aluminium present in 1 mole of potash alum = 54 gso, moles of potash alum prepared by 5.4 g of aluminium = 154∗5.4=0.1mole.

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?

The correct option is B $0.1$ mole
We have,
molar mass of potash alum = 474.3 g/mol
mass of aluminium present in 1 mole of potash alum = 54 g
so, moles of potash alum prepared by 5.4 g of aluminium = $\frac{1}{54} * 5.4 = 0.1 m o l e$ .