If 0.3 mole Zn are added to HCl containing 0.52 mole of HCl, how many moles of $H_{2}$ are produced, according to the reaction
$Z n (s) + 2 H C l (a) \to Z n C l_{2} (a) + H_{2} (g)$

Question

If 0.3 mole Zn are added to HCl containing 0.52 mole of HCl, how many moles of H 2 are produced, according to the reaction Z n ( s ) + 2 H C l ( a ) → Z n C l 2 ( a ) + H 2 ( g )

TestHub · Accepted Answer

Z n ( s ) + 2 H C l ( a ) → Z n C l 2 ( a ) + H 2 ( g ) As we have 0.30 mole of Zn, therefore, HCl will react completely, i.e., HCl is the limiting reactant. 2 moles of HCl produce H 2 = 1 mole So 0.52 mole of HCl will produce H 2 = 1 2 × 0.52 mole = 0.26 mole

Chemistry - Mole Concept Question with Solution | TestHub

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