The reaction of 3.30 g of aluminum with excess of HCl produced 10.5 g of AlCl3. What is the percent yield for the AlCl3? 2Al(s) + 6HCl(aq) ? 2AlCl3(aq) + 3H2(g) 64.4% 20.0% 16.3% 17.1% 35.5%
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The balanced chemical equation for the reaction of aluminum with hydrochloric acid to produce aluminum chloride and hydrogen gas is: \[ 2Al(s) + 6HCl(aq) \rightarrow 2AlCl_3(aq) + 3H_2(g) \] Show more…
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For the reaction: 2 Al (s) + 6 HCl (g) → 2 AlCl3 (s) + 3 H2 (g) 8. Calculate the moles of aluminum required theoretically to produce 854 g of aluminum chloride. 9. What mass of hydrogen can be produced theoretically from 164 g of hydrogen chloride? 10. What is the percent yield of this reaction if 48.8 g of hydrogen are actually produced from 539 g of aluminum?
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2Al(s)+6Hcl(aq)-----> 2AlCl3(s) + 3H2(g) Molar mass Al= 26.98 g/mol Molar mass HCl= 36.45g/mol Molar mass AlCl3=133.33 g/mol Molar mass H2= 2.00g/mol If 15.0 grams of aluminum completely react according to the provided equation and 1.54 grams of hydrogen are collected, what is the percent yield? Write in the CORRECT NUMBER OF SIGNIFICANT FIGURES. -------%
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