lalimares1038
29.03.2021 •
Chemistry
For the reaction 2 Al + 3 H2SO4⟶ 3 H2 + H2(SO4)3 how many grams of sulfuric acid, H2SO4, are needed to react completely with 53.7 g of aluminum, Al??
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Ответ:
292.4 g
Explanation:
2Al + 3H₂SO₄ → 3H₂ + Al₂(SO₄)₃First we convert 53.7 g of Al into moles, using its molar mass:
53.7 g ÷ 27 g/mol = 1.99 mol AlThen we convert Al moles into H₂SO₄ moles, using the stoichiometric coefficients:
1.99 mol Al * = 2.98 mol H₂SO₄Finally we convert 2.98 moles of sulfuric acid into grams, using its molar mass:
2.98 mol H₂SO₄ * 98 g/mol = 292.4 gОтвет:
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Explanation: