CHEM: A 0.250-g sample of a magnesium-aluminum alloy dissolves completely in an

Question

CHEM:

A 0.250-g sample of a magnesium-aluminum alloy dissolves completely in an excess of HCl(aq). When the liberated H_2(g) is collected over water at 29 degree C and 752 torr, the volume is found to be 345 mL. The vapor pressure of water at 29 degree C is 30.0 torr. Now if x is the initial mass of magnesium and y is the initial mass of aluminum, the total number of moles of hydrogen gas produced, n, can be expressed as n = 0.0411x + 0.0556y. If the total mass of the two metals is 0.250 g, then x + y = 0.250, and therefore x = 0.250 - y. Substitute for x in the first equation to express n in terms of y: n = 0.0411(0.250 - y) + 0.0556y The value of n can be found using the ideal gas law. Taking into account the vapor pressure of water, how many moles of hydrogen gas, n, are present in 345 mL at 752 torr and 29 degree C? The value of the gas constant R is 0.08206 L middot atm/(mol middot K). You may also find the conversion factors 1 atm = 760 torr and T_K = T_C + 273 useful. Express the number of moles to three significant figures.

Explanation / Answer

Given-

P = 752torr

Volume of H2 = 345 ml = 0.345L

Vapor pressure of water at 290C =30

T= 29 + 273 = 302K

R = 0.08206 L atm / mol K)

Pressure of the dry H2 = Total pressure - Vapor pressure of water

                                        = 752 torr – 30 torr = 722 torr

Use PV = n RT to calculate moles of H2

(722torr /760torr) (0.345 L) = n (0.0821 L atm / mol K) (302 K)

n = [(722 / 760) (0.345 L)]/[ (0.08206 L atm / mol K) (302 K)]

n = 0.0132 mol (3 sig. figure)

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