A mass of $0.65 g$ of zinc reacts completely with $H C l$ to produce hydrogen gas and a zinc chloride compound. The mass of the chlorine in the compound is $0.71 g$ , what is the empirical formula of this compound? (At. mass of $Z n = 65 g / m o l$ )

Z n C l

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Z n {C l}_{2}

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Z n {C l}_{5}

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Z n {C l}_{4}

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Solution

The correct option is B $Z n {C l}_{2}$
$Z n + H C l \to Z n_{x} C l_{y} + H_{2}$
Given, mass of zinc in $Z n_{x} C l_{y} = 0.659 g$
Mass of chlorine in $Z n_{x} C l_{y} = 0.71 g$
No. of moles of zinc $= \frac{0.65}{65} = 0.01 m o l e s$
No. of moles of chlorine $= \frac{0.72}{35.45} = 0.02 m o l e s$
Ratio of zinc to chlorine $= Z n : C l = 1 : 2$
So, empirical formula $= Z n C l_{2}$

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A mass of 0.65 g of zinc reacts completely with HCl to produce hydrogen gas and a zinc chloride compound. The mass of the chlorine in the compound is 0.71 g, what is the empirical formula of this compound? (At. mass of Zn=65 g/mol)

The correct option is B ZnCl2Zn + HCl →ZnxCly + H2Given, mass of zinc in ZnxCly=0.659 g Mass of chlorine in ZnxCly=0.71 g No. of moles of zinc =0.6565=0.01 molesNo. of moles of chlorine =0.7235.45=0.02 molesRatio of zinc to chlorine =Zn:Cl=1:2So, empirical formula =ZnCl2

A mass of $0.65 g$ of zinc reacts completely with $H C l$ to produce hydrogen gas and a zinc chloride compound. The mass of the chlorine in the compound is $0.71 g$ , what is the empirical formula of this compound? (At. mass of $Z n = 65 g / m o l$ )