2Al + 3H2SO4 -> Al2(SO4)3 + 3H2How many grams of aluminum sulfate would be formed if 250g H2SO4 completely reacted with aluminum?

290.82g

Explanation:

The equation for the reaction is given below:

2Al + 3H2SO4 -> Al2(SO4)3 + 3H2 now, let us obtain the masses of H2SO4 and Al2(SO4)3 from the balanced equation. This is illustrated below:

Molar Mass of H2SO4 = (2x1) + 32 + (16x4) = 2 + 32 +64 = 98g/mol

Mass of H2SO4 from the balanced equation = 3 x 98 = 294g

Molar Mass of Al2(SO4)3 = (2x27) + 3[32 + (16x4)]

= 54 + 3[32 + 64]

= 54 + 3[96] = 54 + 288 = 342g

Now, we can obtain the mass of aluminium sulphate formed by doing the following:

From the equation above:

294g of H2SO4 produced 342g of Al2(SO4)3.

Therefore, 250g of H2SO4 will produce = (250 x 342)/294 = 290.82g of Al(SO4)3

Therefore, 290.82g of aluminium sulphate (Al(SO4)3) is formed.

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