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If 20 g of Na2SO4 is reacted with 20 g of Al(NO3)3 according to the following...

If 20 g of Na2SO4 is reacted with 20 g of Al(NO3)3 according to the following equation:

3 Na2SO4 + 2 Al(NO3)3 --> Al2(SO4)3   + 6 NaNO3

How many grams of the excess reactant will remain after the reaction, assuming that all of the limiting reactant reacts to form the products?

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