Question

What is the maximum mass of s8 that can be produced by combining 89.0 g of each reactant?

Answer 1

Answer : We can produce 125.7 g of
`S_(8)`.

Explanation : The reaction will be

`8SO_(2) + 16H_(2)S -----> 3S_(8) + 16H_(2)O`

The molecular mass of
`SO_(2)` is 64.1 g/mol

and molecular mass of
`H_(2)S` is 34.1 g/mol

For every mole of
`SO_(2)` we would need twice of
`H_(2)S` moles, so for every 3 moles of
`S_(8)` we need 16 moles of
`H_(2)S`

Now, we can calculate number of moles
`S_(8)`

2.61 X (3/16) = 0.49 moles

Here, the molecular mass of
`S_(8)` is 256.8 g

multiplying it with the number of 0.49 moles we get, 256.8 X 0.49 = 125.7 g of
`S_(8)`.

Hence, 125.7 g of
`S_(8)` will be produced.