The reaction of aluminum atoms displacing hydrogen ions from sulphuric acid can be represented as:

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

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

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

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

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

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

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

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

Open in App

Solution

The correct option is A $2 A l + 3 H_{2} S O_{4} \to A l_{2} (S O_{4})_{3} + 3 H_{2}$
The reaction of aluminium atoms displacing hydrogen ions from sulphuric acid can be represented as :
$2 A l + 3 H_{2} S O_{4} \to A l_{2} (S O_{4})_{3} + 3 H_{2}$
This is the balanced chemical equation.
The products of the reaction are aluminium sulphate and hydrogen.

Suggest Corrections

Similar questions

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

X

A l + N a O H \to N a_{3} A l O_{3} + H_{2}

2 A l + 2 O H^{-} + 6 H_{2} O \to 2 [A l {(O H)}_{4}]^{-} + 3 H_{2}

A l + C o n c . H N O_{3} \to A l {(N O_{3})}_{3} + N H_{4} N O_{3} + H_{2} O

K M n O_{4} + H_{2} S O_{4} + H C l \to K_{2} S O_{4} + M n S O_{4} + C l_{2} + H_{2} O

2 A l + 6 H C l \to 2 A l C l_{3} + 3 H_{2}

A l C l_{3}

M n O_{2}

2 N a C l + M n O_{2} + 3 H_{2} S O_{4} \to 2 N a H S O_{4} + M n S O_{4} + C l_{2} + 2 H_{2} O

Join BYJU'S Learning Program