1.

In the reaction, 2Al_(s) + 6hCl_(aq) to 2Al_(aq)^(3+) + 6Cl_(aq)^(-) + 3H_(2)(g)

Answer»

11.2 L `H_(2)(g)` at STP is produced for every mole `HCl_(aq)` consumed
6 L `HCl_(aq)` is consumed for every 3 L `H_(2)(g)` produced
33.6 L `H_(2)(g)` is produced at STP for every mole Al that reacts.
67.2 L `H_(2)`(g) at STP is produced for every mole Al that reacts.

Solution :`2Al_(s) + 6HCl_(aq) to 2Al_(aq)^(3+) + 6Cl_(aq)^(-) + 3H_(2)(g)`
At STP, 6 moles of HCl produce 3 moles of `H_(2) = 3 xx 22.4 L` of `H_(2)`
Against at STP, 2 moles of Al produce 3 moles of `H_(2) = 3 xx 22.4` L of `H_(2)`
`THEREFORE 1` mole of Al produce `=(3 xx 22.4)/2 = 33.6` L of `H_(2)` at STP.


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