How many ml of 6.00m sulfuric acid are needed to react

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Reference no: EM13699859

Problem- How many mL of 6.00M sulfuric acid are needed to react with 17.6g of aluminum according to the reaction below?

2Al(s) + 3H2SO4(aq) ----> Al2(SO4)3(aq) +3H2(g)

So far I have 17.6g Aluminum * 1mole Al/ 26.9815

Not sure how to convert from Moles to Liters/mL

Anyone can tell me how to solve it step by step so that I will be able to understand what is going on?

Reference no: EM13699859

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