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?