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Considered the balanced reaction, what mass of aluminum must react to produce 0.93 L of H2(g) at STP? 2H3PO4(aq) + 2Al(s) —> 2AlPO4(aq) + 3H2(g)

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0.93L H2 x 1 mole/22.4 L x 2 mole Al/3 mole H2 x 26.98 g Al/1 mole Al
= 374 g Al
User NatGordon
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