For the equation 3 H2SO4 + 2 Al - > 3 H2 + Al2 (SO4) 3, suppose you wish to prepare 28.7 grams of H2. How many grams of H2SO4 and of Al would you require so that all of the H2SO4 and Al were combined?

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Chemistry

Houston

28 November, 00:54

For the equation 3 H2SO4 + 2 Al - > 3 H2 + Al2 (SO4) 3, suppose you wish to prepare 28.7 grams of H2. How many grams of H2SO4 and of Al would you require so that all of the H2SO4 and Al were combined?

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Madden Oconnell · Answer 1

In your case, the reaction produced 28.7g * (1 mole H2/2.016g) = 14.236 moles H2 This implies that the reaction consumed 14.236 moles H2 * (2 moles Al / 3 moles H2) = 9.491 moles Al Finally, to convert the number of moles of each reactant to moles, use the molar masses of the two compounds. 9.491 moles Al * (26.982 g/1 mole Al) = 256 g The answers are rounded to three sig figs, the number of sig figs you have for the mass of hydrogen gas.