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2Al(s) + 6HCl(aq) 2AICI3(aq) + 3H(g) According to the equation above, how many grams of aluminum are needed to completely react with 3.83 mol of

2Al(s) + 6HCl(aq) 2AICI3(aq) + 3H(g) According to the equation above, how many grams of aluminum are needed to completely react with 3.83 mol of hydrochloric acid? Al= 26.981 g/mol H= 1.008 g/lol Cl= 35.45 g/mol Answer format one significant figure: OE00 g

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