Question

Find out the amount of Al2(SO4)3 which form when 11:5 g Al reacts with H2SO4 conc. in reaction​

Answer 1

Answer:

sorry I don't know the answer

Answer 2

Given:

11.5g of Al reacts with H₂SO₄(conc.)

To find:

Amount of Al₂(SO₄)₃

Solution:

The balanced chemical reaction is presented below-

2Al  + 3H₂SO₄ →  Al₂(SO₄)₃ + 3H₂

firstly we have to find the limiting agent,

so, 2×27g of Al react = 3×98g of H₂SO₄

    11.5g of Al react = $\left( {\frac{{3 \cdot 98}}{{2 \cdot 27}}} \right) \cdot 11.5$g of H₂SO₄

                                =62.61g

that means, Al is a limiting agent.

To find the amount of Al₂(SO₄)₃,

2×27g of Al gives Al₂(SO₄)₃ = 342g

11.5g of Al gives Al₂(SO₄)₃ = 72.83g

So, the amount of Al₂(SO₄)₃ formed is 72.83g