2Al + 6HCl ? 2AlCl3 + 3H2 If 3.0 grams of H2 were produced, how many grams of Al reacted?
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Multiply the moles of each reactant by the mole ratio between the reactant and AlCl3 in the balanced equation, so that the moles of each reactant cancel, leaving moles of AlCl3. Cl2 is the limiting reactant, therefore a maximum of 5.3 mol AlCl3 can be produced.
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