- When 1.00 mol Zinc reacts with excess hydrochloric acid, \( 1.00 \mathrm{~mol} \mathrm{H}_{2} \) is produced. At \( 25^{\circ} \mathrm{C} \) and 1.00 atm , this amount of \( \mathrm{H}_{2} \) occupies 24.0L. Calculate the work done by the chemical system in pushing back the atmosphere.
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The reaction is between zinc and hydrochloric acid, producing hydrogen gas. The conditions are 25°C and 1.00 atm, with 1.00 mol of \( \mathrm{H}_2 \) produced, occupying 24.0 L. Show more…
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Zinc metal reacts with excess hydrochloric acid to produce hydrogen gas: Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) In one experiment, a sample of Zn reacts, and the gas produced is collected by water displacement. The gas sample has a temperature of 22.00 °C, a volume of 645.0 mL, and a pressure of 742.0 mm Hg. Calculate the amount (in moles) of hydrogen gas produced in the reaction. The vapor pressure of water is 19.83 mm Hg at 22.00 °C. _______mol
Adi S.
a. Calculate the amount of work done against an atmospheric pressure of 1.00 atm when 500.0 g of zinc dissolves in excess acid at 30.0°C. Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g) Assume the volume of reactants is negligible compared to that of the vapor produced. b. If the ΔH° for this reaction is -1177 kJ, what is ΔE for this reaction?
Shaiju T.
Calculate $\Delta H$ for the reaction $$ \mathrm{Zn}(\mathrm{s})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{ZnO}(\mathrm{s}) \quad \Delta H=? $$ given the equations $$ \begin{aligned} \mathrm{Zn}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{ZnCl}_{2}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{~g}) & \Delta H=-152.4 \mathrm{~kJ} \\ \mathrm{ZnO}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{ZnCl}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \\ \Delta H=-90.2 \mathrm{~kJ} \\ 2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\ell) & \Delta H=-571.6 \mathrm{~kJ} \end{aligned} $$
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