Small quantities of hydrogen gas can be prepared in the laboratory by the additi

Question

Small quantities of hydrogen gas can be prepared in the laboratory by the addition of aqueous hydrochloric acid to metallic zinc. Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g) Typically, the hydrogen gas is bubbled through water for collection and becomes saturated with water vapor. Suppose 360. mL of hydrogen gas and water vapor is collected at 30. °C and has a total pressure of 1.470 atm. The vapor pressure of water is 32 torr at 30. °C. What is the partial pressure of hydrogen gas in the sample? How many moles of hydrogen gas are present in the sample? How many grams of zinc must have reacted to produce this quantity of hydrogen?

_______atm H2

________mol H2

__________g Zn

2.

A 20.0-L nickel container was charged with 0.809 atm of xenon gas and 1.37 atm of fluorine gas at 400°C. The xenon and fluorine react to form xenon tetrafluoride. What mass (g) of xenon tetrafluoride can be produced assuming 100% yield?

_______ g XeF4

Explanation / Answer

Zn(s) + 2HCl (aq) ---> ZnCl2 ( aq) + H2 (g) ,

HCl volume = 360 ml = 0.36 L , P H2 = ptotal - PH2O = 1.47 - ( 32/760) = 1.4279 atm ( 32 torr = 32/760 atm)

we use PV = nRT to get moles of H2 , T = 30C = 30+273 = 303 K

1.4279 x 0.36 = n x 0.08206 x 303

n = 0.020674 = moles of H2

pH2 = 4.7219 atm

moles of Zn = moles of H2 = 0.020674

mass of Zn = moles x molar mass of Zn = 0.020674 x 65.38 = 1.35 g

2) X2 + 2F2 ---> XeF4    is the balanced eq

we use PV = nRT equation to find moles of Xe

0.809 x 20 = n x 0.08206 x 673               ( T = 400C = 400+273 = 673 K)

n = 0.293 moles of Xe

1.37 x 20 = n x 0.08206 x 673

n = 0.49614 moles of F2

as per reaction 1 Xe reacts with 2F2 , hence 0.293 Xe needs 2 x0.293 = 0.586 moles F2 but we had only 0.49614 moles F2 , hence F2 is limiting reagent while Xe is excess reagent

hence XeF4 moles = (1/2) F2 moles = ( 1/2) x 0.49614 = 0.24807

XeF4 mass = moles x molar mass of XeF4 = 0.24807 x 207.2836 = 51.42 g

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