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Balanced equation first.

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

( only onfo on H2SO4 and it does say; ' completely reacted. ' So, H2SO4 limits and drives the reaction )( if it did not say this you would have to find mol Al and see what limits )

250 grams H2SO4 (1mol H2SO4/98.086g )(1mol Al2(SO4)3/3mol H2SO4 )(357.17g/1mol Al2(SO4)3 )

= 303 grams of Al2(SO4)3

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291 grams

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Q: How many grams of aluminum sulfate would be formed if 250g H2 SO4 completely reacted with aluminum?
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