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How many grams of zinc chloride could be formed from the reaction of 4.96g of zinc with excess HCl?
Wiki User
∙ 9y ago
The reaction of zinc with hydrochloric acid is simplified as Zn + 2 HCl = ZnCl2 + H2. This means for every mole of zing, 1 mole of ZnCl2 is formed. 4.96 grams zinc is equal to .0759 moles, so 5.17 grams of zinc chloride would be optimally produced.
Wiki User
∙ 9y ago
Wiki User
∙ 9y agoZinc reacts with two parts of Hydrogen Chloride to form Zinc Chloride and Hydrogen. n(Zn)=9.76/65.37=0.1493. Similarly n(ZnCl2)/n(Zn)=1, therefore n(ZnCl2) is 0.1493 resulting in 0.1493(65.37+2*35.45)= 20.3grams of ZnCl2.
Wiki User
∙ 15y agoI hope this makes sense. First write out a balanced equation.
Zn + 2HCl - ZnCl2 + H2
since you are told that the HCl is in excess, you then assume that the limiting reaagent is Zn. given that the mole to mole ratio of Zn to Zncl2 is 1:1, and that the number of moles of zinc is 0.149(9.76/65.38), it can easily be deduced that the number of moles in the ZnCl2 sample is also 0.149(since their mole to mole ratio is 1:1). Then convert the number of moles to grams by multiplying 0.149 by the molar mass of ZnCl2(65.38+(35.5*2). This will give 20.32g.
If anyone has a contrary opinion please feel free to put in your suggestions and correct me and evryone else because learning is a continuous process.
Thank you.
Wiki User
∙ 13y agoZn + 2HCl -> ZnCl2 + H2
4 grams zinc (1 mole Zn/65.41 grams)(1 mole H2/1 mole Zn)
= 0.0621 moles hydrogen produced
I assume this reaction is at STP; so PV =nRT
(1 atm)(volume) = (0.0621 mole H2)(0.08206 L*atm/mol*K)(298.15 K)
= 1.519 Liters
centimeters cubed is the same as milliliters, so....
1519 cm^3
Wiki User
∙ 15y ago10.378 g
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