How much volume of chlorine is required to prepare 89.6 L of HCI gas a

1.	How much volume of chlorine is required to prepare 89.6 L of HCI gas at STP?
Answer» Solution :`UNDERSET(1xx22.4L)(H_(2(g))) + underset(1xx22.4L)(Cl_(2(g))) rarr underset(2xx22.4L)(2HCl_(g))` 2 `XX` 22.4 L of HCI is produced by 22.4 L of `Cl_(2)` `:.`89.6 L of HCI will be produced by `22.4/(2xx22.4) xx 89.6L` = `44.8` L of chlorine.

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