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.