221 grams is the mass of oxygen gas is consumed in a reaction that produces 4.60 mol sulfur dioxide
The given question is incomplete as it lacks a chemical reaction, however, the chemical reaction is :
Hydrogen sulfide gas burns in oxygen to produce sulfur dioxide and water vapor.
[tex]2H_2S(g)+3O_2(g)\rightarrow 2SO_2(g)+2H_2O(g)[/tex]
Explanation:
Given:
The chemical reaction:
[tex]2H_2S(g)+3O_2(g)\rightarrow 2SO_2(g)+2H_2O(g)[/tex]
To find:
Mass of oxygen gas consumed in a reaction to produce 4.60 moles of sulfur dioxide gas.
Solution:
The moles of sulfur dioxide gas produced =4.60 mol
[tex]2H_2S(g)+3O_2(g)\rightarrow 2SO_2(g)+2H_2O(g)[/tex]
According to a chemical reaction, 2 moles sulfur dioxide is obtained from 3 moles of oxygen gas, then 4.60 moles of sulfur dioxide will be obtained from:
[tex]=\frac{3}{2}\times 4.60mol=6.9\text{ mol of }O_2[/tex]
Mass of 6.9 moles of oxygen gas:
[tex]6.9 mol\times 31.998 g/mol=220.79 g\approx 221 g[/tex]
221 grams is the mass of oxygen gas is consumed in a reaction that produces 4.60 mol sulfur dioxide.
Learn more about the unitary method and moles here:
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