Al + H_{2}SO_{4} ==> Al_{2}(SO_{4})_{3} + H_{2}

What mass of aluminum is needed to produce 0.766 moles of hydrogen?

If 49.3 g of hydrogen sulfate is reacted, how many moles of aluminum is also reacted?

First, the reaction equation has to be balanced:

2 Al + 3 H2SO4 ==> Al2(SO4)3 +3 H2

So the molar ratio between Al and H2 is 2:3, so to produce 0.766 moles of H2, 0.766x2/3=0.511 moles of Al.

The molar mass of Al is 27 g/mol, so the mass of Al needed=27x0.511=13.8 g.

The molar mass of H2SO4=2x1+32+16x4=98 g/mol

So the moles of H2SO4=49.3/98=0.503 mole

The molar ratio between Al and H2SO4 is 2:3.

So when 0.503 moles of H2SO4 is reacted, 0.503x2/3=0.335 moles of Al is also reacted.

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