What is the limiting reactant in the reaction of 12.0 g of so2 with 8.0 g of h2s and their reaction? what mass of sulfur will be produced?

The balanced equation for the reaction is as follows;
2H₂S + SO₂ —> 2H₂O + 3S
Stoichiometry of H₂S to SO₂ is 2:1
Limiting reactant is fully used up in the reaction and amount of product formed depends on amount of limiting reactant present.
Number of H₂S moles - 8.0 g / 34 g/mol = 0.24 mol of H₂S
Number of SO₂ moles = 12.0 g / 64 g/mol = 0.188 mol of SO₂
According to molar ratio of 2:1
If we assume H₂S to be the limiting reactant
2 mol of H₂S reacts with 1 mol of SO₂ 
Therefore 0.24 mol of H₂S requires - 1/2 x 0.24 = 0.12 mol of SO₂ 
But 0.188 mol of SO₂ is present therefore SO₂ is in excess and H₂S is the limiting reactant.
H₂S is the limiting reactant
Amount of S produced depends on amount of H₂S present
Stoichiometry of H₂S to S is 2:3
2 mol of H₂S forms 3 mol of S
Therefore 0.24 mol of H₂S forms - 3/2 x 0.24 mol = 0.36 mol of S
Mass of S produced = 0.36 mol x 32 g/mol = 11.5 g of S is produced

That result from subduction of oceanic plates beneath lighter continental plates. most of the earth's volcanoes are located around the pacific ring of fire because that the location of most of the earth's subduction zones.

m=9

step-by-step explanation:

7m+8=71 subtract   8 from 71 which is 63 than divide it by 7m   equal 9.