Question #105983
Calculate the mass of hydrogen formed when 3.79 g of aluminum reacts with excess hydrochloric acid according to the balanced equation: 2Al + 6 HCl → 2 AlCl3 + 3 H2
molar masses
Al = 26.98 g/mol HCl = 36.46 g/mol H2 = 2.02 g/mol AlCl3 = 133.33 g/mol
molar masses
Al = 26.98 g/mol HCl = 36.46 g/mol H2 = 2.02 g/mol AlCl3 = 133.33 g/mol
Expert's answer
2Al + 6HCl → 2AlCl3 + 3 H2
The mole ratio between aluminium and the hydrogen produced is 2:3
3.79g of Al= 3.79/26.98 =0.1405 moles
If 0.1405 moles of Al reacted then 0.2108 moles of hydrogen were produced.
0.2108 moles of hydrogen =2.02*0.2108=0.43g
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Calculate the mass of hydrogen formed when 25 grams of aluminum reacts with excess hydrochloric acid.