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Sagot :
The first thing to do when solving this problem is to make sure the equation is balanced:
[tex]2H_2S_{(g)}+3O_{2(g)}\rightarrow2H_2O_{(g)}+2SO_{2(g)}[/tex]We want to know the maximum amount of SO2 that can be produced.
Given:
mass of H2S = 1060 g
mass of O2 = 2890 g
We have to find the limiting reagent, and use its molar ratio to find number of moles of SO2, then we can find the mass produced.
H2S:
n = m/M where n is the number of moles, m is the mass and M is the molar mass. Molar mass can be found by adding atomic masses of H2 + S.
n = 1060 g/34,1 g/mol
n = 31.085 mol
O2:
n = m/M
n = 2890 g/31.998
n = 90.318 mol
So Oxygen has a higher number of moles, so it is reactant in excess and H2S the limiting reactant/reagent since it has less number of moles.
We can use the stoichiometry to find the molar ratios of H2S:SO2. The molar ratio is 2:2.
Therefore the number of moles of SO2 = 31.085 x (2/2) = 31.085 mol
From the number of moles, we can find the mass of SO2.
m = n x M
m = 31.085 mol x 64,066 g/mol
m = 1991.5 g of sulfur dioxide will be produced.
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