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First we will need to the complete chemical reaction involving sulfuric acid and alluminium.
2Al + 3H2SO4 -----> Al2(SO4)3 + 3H2
From the reaction
2 moles of Al react with 3 moles of H2SO4 [ That is is 2:3 ratio]
18 moles of Al will react with 27 moles of H2SO4.
(18 * 3)/2 = 27 moles of H2SO4.
Therefore 18 moles of Al will require 27 moles of H2SO4 to react.
Aluminum reacts with `H_2SO_4` to form `Al_2(SO_4)_3` . The equation of the chemical reaction is:
`2Al + 3H_2SO_4 -> Al_2(SO_4)_3 + 3H_2 `
2 moles of aluminum react with 3 moles of hydrogen sulfide. 18 moles of Al will react with (18/2)*3 = 27 moles of hydrogen sulfide.
18 moles of aluminum react with 27 moles of hydrogen sulfide in the reaction involving the two.
You should write the chemical reaction and you need to balance the equation such that:
Al + H_2SO_4 -> Al_2(SO_4)_3 + H_2
You need to balance the group SO_4, hence, you should multiply by 3 to the left such that:
Al + 3H_2SO_4 -> Al_2(SO_4)_3 + H_2
You need to balance Al and H2, hence, you should multiply Al by 2, to the left, and you need to multiply H_2 by 3 such that:
2Al + 3H_2SO_4 -> Al_2(SO_4)_3 + 3H_2
You need to determine the mole ratio of H_2SO_4 to Al, hence, the mole ratio is of 2:3.
2 moles of Al reacts with 3 moles of H_2SO_4
18 moles of Al reacts with x moles of H_2SO_4
Hence, x = 18*3/2 = 9*3 = 27 moles of H_2SO_4
Hence, evaluating the number of moles of H_2SO_4 that reacts with 18 moles of Al yields 27 moles.
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