of according to the following reaction?
2Al+3 H2SO4 yield Al2 (SO4)3+3H2
2Al+3 H2SO4 yield Al2 (SO4)3+3H2
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2 Al + 3 H2SO4 → Al2(SO4)3 + 3 H2
(26.7 g H2) / (2.0159 g H2/mol) x (3/3) / (6.6 mol/L) = 2.01 L H2SO4
(26.7 g H2) / (2.0159 g H2/mol) x (3/3) / (6.6 mol/L) = 2.01 L H2SO4
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:)