Using the stoichiometry that the balanced equation (2 mol Al : 3 mol H2SO4) and also dimensional analysis, us can find the mass of Al required to react v 26.5 ml the 0.542 M H2SO4.

You are watching: Al(s)+h2so4(aq)→al2(so4)3(aq)+h2(g)

How numerous moles H2SO4 room present?

26.5 ml x 1 L/1000 ml x 0.542 mol/L = 0.014363 moles H2SO4

How numerous moles Al are necessary to react v this?

0.014363 mol H2SO4 x 2 mol Al / 3 mol H2SO4 = 0.009575 moles Al

Converting this to grams and then come micrograms (ug), we have...

0.009575 mol Al x 26.98 g Al/mol x 106 ug//g = 2.58x105 ug Al needed


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