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Stoichiometric Calculations

If 241.5 g of...

Question

If 241.5 g of zinc sulphate $(Z n S O_{4})$ reacts with 81 g of aluminium to give aluminium sulphate. Calculate the mass of the excess reagent left.

$3 Z n S O_{4} + 2 A l \to A l_{2} (S O_{4})_{3} + 3 Z n$
(take molar mass of $Z n$ as 65 g)

27 g

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54 g

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81 g

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161 g

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Solution

The correct option is B $54 g$
Given balanced chemical reaction is:
$3 Z n S O_{4} + 2 A l \to A l_{2} (S O_{4})_{3} + 3 Z n$

The molar mass of $Z n S O_{4} = 65 + 32 + 64 = 161 g / m o l$

$So, 241.5 g Z n S O_{4} = \frac{241.5}{161} m o l = 1.5 m o l Z n S O_{4}$
$Again, 81 g A l = \frac{81}{27} m o l = 3 m o l A l$

From the balanced chemical equation,
3 moles of $Z n S O_{4}$ reacts with 2 moles of $A l$ .

Thus, 1.5 moles of $Z n S O_{4}$ will react with 1 mole of $A l$ .

We are given 3 moles of Al.

Thus, the excess reagent here is $A l$ .
$Amount of excess A l = 3 - 1 m o l = 2 \times 27 g = 54 g$

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Similar questions

(Z n S O_{4})

3 Z n S O_{4} + 2 A l \to A l_{2} (S O_{4})_{3} + 3 Z n

Z n

(Z n S O_{4})

3 Z n S O_{4} + 2 A l \to A l_{2} (S O_{4})_{3} + 3 Z n

Z n

A l C l_{3}

H_{2} S O_{4}

A l C l_{3} + 3 N a O H \to A l (O H)_{3} + 3 N a C l

2 A l (O H)_{3} + 3 H_{2} S O_{4} \to A l_{2} (S O_{4})_{3} + 6 H_{2} O

A l C l_{3}

H_{2} S O_{4}

A l C l_{3} + 3 N a O H \to A l (O H)_{3} + 3 N a C l

2 A l (O H)_{3} + 3 H_{2} S O_{4} \to A l_{2} (S O_{4})_{3} + 6 H_{2} O

68.4 g

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

500 m L

(A l^{3 +})

(S O_{4}^{2 -})

A l_{2} (S O_{4})_{3} = 342 g m o l^{- 1})

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