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When `22.4 L` of `H_(2)(g)` is mixed with 11.2 of `Cl_(2)(g)`, each at STP, the moles of HCl(g) formed is equal toA. 1 mole of HCl (g)B. 2 moles of HCl (g)C. 0.5mole of HCl (g)D. 1.5 mole of HCl(g) |
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Answer» Correct Answer - B The given problem is related to the concept of stoichiometry of chemical equations. Thus, we have to convert the given volumes into their moles and then, identity the limiting reagent . (possessing minimum number of moles and gets completely used up in the reaction). The limiting reagent gives the moles of product formed in the reation. `underset(22.4L)(H_(2)(g)) + underset(11.2L)(cl_(2)(g)) to underset("2 mol")(2HCl(g))` `because` 22.4 volume at STP is occupied by `Cl_(2) = 1 "mole"` `therefore ` 11.2L volume will be occupoied by `Cl_(2) = (1 xx 11.2)/(22.4)" mole" = 0.5 mol` Thus `underset(1 mol)(H_(2)(g)) + underset(0.5 mol)(Cl_(2)(g)) to 2HCl (g)` Since `Cl_(2)` possesses minimum number of moles , thus it is the limiting reagent As per equation , `mol Cl_(2) = 2mol` HCl `therefore 0.5 mol Cl_(2) = 2 xx 0.5 mol HCl` `= 1.0 mol HCl` Hence , 1.0 mole of HCl (g) is produced by 0.5 mole of `Cl_(2) (or 11.2L)` |
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