A 0.1164 g Zn-Al sample is dissolved in acid and gives 125.85 mL H2 gas when mea

Question

A 0.1164 g Zn-Al sample is dissolved in acid and gives 125.85 mL H2 gas when measured over water at 22.8°C and 751.9 mm Hg barometric pressure. The vapor pressure of water is 21.0 torr at this temperature. (R = 0.08206 L atm/(K mol))

Calculate:

a. Pressure of H2 gas in atm.

b.Moles H2 gas

c. Moles gas per gram sample (NH2)

d. Percent Al as calculated from the trendline equation. (y = mx + b from the graph) (Hint: answer c. = y, solve for x)

e. Show the relationship of the NH2 value and %Al on your graph as the intersection of a horizontal and vertical line at a point on the line. (You can hand-draw these in INK using a ruler)

Explanation / Answer

Mols of gas produced
(0.989 * 0.126)/(0.082*295.8)= 0.0051 mols wet gas
(751.9-22.8)/751.9 = mol fraction of dry gas = 0.97 dry gas 0.005 mol
write two equations for aluminum +HCL and zinc + HCL
2AL = 3 H2 Zn = H2. Each AL makes 1.5 H2 (mols)
Set up an equation in X fraction of zinc in sample and 0.1164- X as fraction of AL in sample.Convert to mols H2 for each metal. Add and equal total to 0.005 mols of H2. I caculated X = 0.3

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