1.

For the formation of 3.65 g of hydrogen chloride gas, what volume of hydrogen and chlorine gas are required to N.T.P. conditions?

Answer»

1.12L, 1.12L
1.12L, 2.24L
3.65 L, 1.83 L
1L, 1L

Solution :`underset("22.4 L at NTP")(H_(2)(g))+underset("22.4 L at NTP")(Cl_(2)(g))to underset(2xx36.5g)(2HCL(g))`
To prepare `2xx36.5 g` of HCL, `H_(2)` and `Cl_(2)` REQUIRED at N.T.P. are 22.4 L each
`:.` To prepare 3.65 g of HCl, `H_(2)` and `Cl_(2)`
required at N.T.P. are `= (22.4)/(2xx36.5)xx3.65`
`=1.12L`
Thus, 1.12 L of `H_(2)` and 1.12 L of `CL_(2)` are required


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