This is the question. Zinc metal reacts with hydrochloric acid according to thef

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This is the question.
Zinc metal reacts with hydrochloric acid according to thefollowing balanced equation.    Zn (s) + 2HCl (aq) arrow ZnCl2 (aq) + H2 (g) When 0.103 g of Zn (s) is combined with enough HCl to make50.0 mL of solution in a coffee-cup calorimeter, all of the zincreacts, raising the temperature of the solution from 22.5 degrees Cto 23.7 degrees C. Find the change in H rxn for this reactionas written. (Use 1.0 g/mL for the density of the solution and4.18 J/g x degrees C as the specific heat capacity.)
Thank you so much for helping me and please can you list thesteps so I can understand good?
Zinc metal reacts with hydrochloric acid according to thefollowing balanced equation.    Zn (s) + 2HCl (aq) arrow ZnCl2 (aq) + H2 (g) When 0.103 g of Zn (s) is combined with enough HCl to make50.0 mL of solution in a coffee-cup calorimeter, all of the zincreacts, raising the temperature of the solution from 22.5 degrees Cto 23.7 degrees C. Find the change in H rxn for this reactionas written. (Use 1.0 g/mL for the density of the solution and4.18 J/g x degrees C as the specific heat capacity.)
Thank you so much for helping me and please can you list thesteps so I can understand good?

Explanation / Answer

We Know that :      The Reaction is :         Zn ( s ) + 2 HCl --------> ZnCl2 + H2       The weight of the Zntaken = 0.103 g        Number of moles of Zn taken = 0.103 g / 65.3 g / mol                                                        = 0.001577 mol         The volume of HCl taken =  50.0 mL         The density of thesolution = 50.0 mL x 1.0 g / mL                                                   = 50.0 g           Numberof moles of HCl  = 1.369 mol              In the above Reaction Zn acts as Limiting Reagent.           Thenumber of moles of ZnCl2   formed = 0.001577mol          According to Thermodynamical Relationship :              Q = m x s x t                    = 0.001577 mol x 136.4 g / mol x 4.18 J /g 0 C x ( 23.7 - 22.5)0C                     = 1.0789 J         The weight of the Zntaken = 0.103 g        Number of moles of Zn taken = 0.103 g / 65.3 g / mol                                                        = 0.001577 mol         The volume of HCl taken =  50.0 mL         The density of thesolution = 50.0 mL x 1.0 g / mL                                                   = 50.0 g           Numberof moles of HCl  = 1.369 mol              In the above Reaction Zn acts as Limiting Reagent.           Thenumber of moles of ZnCl2   formed = 0.001577mol          According to Thermodynamical Relationship :              Q = m x s x t                    = 0.001577 mol x 136.4 g / mol x 4.18 J /g 0 C x ( 23.7 - 22.5)0C                     = 1.0789 J  

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