One way of preparing H2 (g) is given by the reaction of Zn (s) with HCl (aq).
Zn (s) + 2HCl (aq) <=> H2 (g) +ZnCl2 (aq).
The reaction is exothermic and the equilibrium constant is 3.8E5 (3.8 X 10^5) at room temperature (298K). Assuming that the reaction is at equilibrium what are the effects of the following changes on the equilibrium concentration of H2 (g). Explain based on the reaction quotient compared to the equilibrium constant at the point the change is made.
a) Increasing the concentration of HCl (aq) from 0.1 M to 0.2 M.
b) Adding 0.01 M NaOH to the reaction at equilibrium
c) Heating the equilibrium mixture to 450K. Will this equilibrium constant change?
One way of preparing H2 (g) is given by the reaction of Zn (s) with HCl...
Zinc reacts with hydrochloric acid according to the reaction equation Zn(s)+2HCl(aq)⟶ZnCl2(aq)+H2(g)Zn(s)+2HCl(aq)⟶ZnCl2(aq)+H2(g) How many milliliters of 2.00 M HCl(aq)2.00 M HCl(aq) are required to react with 8.55 g Zn(s)?
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Zinc reacts with hydrochloric acid according to the reaction equation shown. Zn(s)+2HCl(aq)⟶ZnCl2(aq)+H2(g) How many milliliters of 5.50 M HCl(aq) are required to react with 2.95 g of an ore containing 43.0 % Zn(s) by mass?
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In the following equation, what is oxidized? Zn(s) + 2HCl(aq) ---> ZnCl2(aq) + H2(g) Group of answer choices Zinc Hydrogen in HCl Chlorine in HCl This is not an oxidation-reduction reaction
8) Balance the equation for the following reaction. Zn(s) + HCl(1) → ZnCl2(aq) + H2(g) -
Hydrogen gas can be prepared by reaction of zinc metal with aqueous HCl: Zn(s)+2HCl(aq)→ZnCl2(aq)+H2(g) How many grams of zinc would you start with if you wanted to prepare 5.05 L of H2 at 260 mm Hg and 26.0 Celsius?
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