.) This reaction occurs in large test tube at 32.0 ° C: 2 Al (s) +3 H2SO4 (aq) → 6 H2 (g) + Al2 (SO4) 3 (aq) A tube going through the stopper of the test tube feeds into a graduated cylinder that has been inverted in a tub of water. At the end of the reaction, the total pressure of gas in the inverted cylinder is 435. torr, and the volume of gas collected is 455. mL. What mass of aluminum must have reacted?

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1.) This reaction occurs in large test tube at 32.0 ° C: 2 Al (s) +3 H2SO4 (aq) → 6 H2 (g) + Al2 (SO4) 3 (aq) A tube going through the stopper of the test tube feeds into a graduated cylinder that has been inverted in a tub of water. At the end of the reaction, the total pressure of gas in the inverted cylinder is 435. torr, and the volume of gas collected is 455. mL. What mass of aluminum must have reacted?
Expert Solution
Introduction

Reaction:
2 Al (s) + 3 H2SO4 (aq) →  6 H2 (g) + Al2(SO4)3 (aq)

Use an ideal gas equation to solve for the moles of H2 gas evolved,
PV = nRT 
n = PV/RT 

Given that,
Total pressure of the gas = P = 435 torr = (435torr / 760torr) x 1atm = 0.5724 atm 
Volume of the gas collected = V = 455mL = 0.455 L 
Temperature = 32oC = 32+273.15 = 305.15 K 
R = 0.082057 L atm mol-1K-1 

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