When aluminum is placed in concentrated hydrochloric acid hydrogen gas is produced.
2AL(s) + 6HCI (aq) ---> 2AlCl3(aq) +3H2(g)
What volume of H2(g) is produced when 7.80 g of Al (s) reacts at STP
2AL(s) + 6HCI (aq) ---> 2AlCl3(aq) +3H2(g)
What volume of H2(g) is produced when 7.80 g of Al (s) reacts at STP
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Things to have in mind are
1. One mole = 22.4L of gas
2. Mole ratio of Al to H2 is 2:3
7.8g mole * 1mole/27g Al tells us how many moles of Al we have in this case, 0.28888
Since the mole ratio is 2:3,
0.2888 : x = 2:3
x 0.433333 = number of H2 mole we have
since 1 mole = 22.4
0.43333 mole = 9.70666666 Liters of gas
1. One mole = 22.4L of gas
2. Mole ratio of Al to H2 is 2:3
7.8g mole * 1mole/27g Al tells us how many moles of Al we have in this case, 0.28888
Since the mole ratio is 2:3,
0.2888 : x = 2:3
x 0.433333 = number of H2 mole we have
since 1 mole = 22.4
0.43333 mole = 9.70666666 Liters of gas