Zinc and magnesium react with hydrochloric acid to produce themetal chlorides an

Question

Zinc and magnesium react with hydrochloric acid to produce themetal chlorides and hydrgoen gas. A 10.00g mixture of Zn and Mg isreacted with the stoichiometric quantity of HCl. The reactionmixture is then reacted with 156ml of 3.00M silver nitrate toproduce the maximum possible quantity of silver chloride. First, determine the % magnesium in the mixture. Then, if 78.0ml of HCl was added, what was the molarity of theHCl? Zinc and magnesium react with hydrochloric acid to produce themetal chlorides and hydrgoen gas. A 10.00g mixture of Zn and Mg isreacted with the stoichiometric quantity of HCl. The reactionmixture is then reacted with 156ml of 3.00M silver nitrate toproduce the maximum possible quantity of silver chloride. First, determine the % magnesium in the mixture. Then, if 78.0ml of HCl was added, what was the molarity of theHCl?

Explanation / Answer

Let x is mole of Zn        y is mole of Mg Mass of Zn: 65.39x(g) Mass of Mg: 24.31y(g) We have : 65.39x +24.31y = 10(1) Reaction: Zn + 2HCl --> ZnCl2 +H2                 x          2x           x                 x          2x           x                 Mg+ 2HCl --> MgCl2 +H2                y           2y             y                y           2y             y Mole of AgNO3: n =CM*V = 3.00M *0.156L = 0.468(mol) Given: produce the maximum possible quantityof silver chloride => assume this is completed reactions ZnCl2 + 2AgNO3 -->Zn(NO3)2 + 2AgCl x                                                      2x MgCl2 + 2AgNO3 --->Mg(NO3)2 + 2AgCl y                                                         2y Mole of AgCl total: 2x +2y = 0.468 (2) From (1) and (2) => solve for x and y => x =0.105(mole)                                                                y= 0.129(mole) = mole of Mg Mass of Mg: 24.31*0.129 = 3.14g % of Mf: 3.14g / 10g *100= 31.4% b)if 78.0ml of HCl was added, what was the molarity of theHCl? From the first reaction, we have total mole of HCl is 2x+2y(mole) Substitute x and y in that => mole of HCl : 2*0.105 +2*0.129 = 0.468(mol) Molarity of HCl: 0.468mol / 0.078L = 6M b)if 78.0ml of HCl was added, what was the molarity of theHCl? From the first reaction, we have total mole of HCl is 2x+2y(mole) Substitute x and y in that => mole of HCl : 2*0.105 +2*0.129 = 0.468(mol) Molarity of HCl: 0.468mol / 0.078L = 6M

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