What is the limiting reagent when 3.10 mol of $S O_{2}$ $SO_2$ reacts with 2.75 mol of $O_{2}$ $O_2$ according to the equation: $S O_{2} + O_{2} \to S O_{3}$ $SO_2 + O_2 -> SO_3$ ?

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What is the limiting reagent when 3.10 mol of $S O_{2}$ $SO_2$ reacts with 2.75 mol of $O_{2}$ $O_2$ according to the equation: $S O_{2} + O_{2} \to S O_{3}$ $SO_2 + O_2 -> SO_3$ ?

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Answer 1 · 2016-06-08T18:54:55

Your equation is incorrect; it is not stoichiometrically balanced.

Explanation:

Synthesis of sulfur trioxide is represented stoichiometrically:

$S O_{2} (g) + \frac{1}{2} O_{2} (g) \to S O_{3} (g)$ $SO_2(g) + 1/2O_2(g) rarr SO_3(g)$

This equation tells us that $64 \cdot g$ $64*g$ of sulfur dioxide reacts with $16$ $16$ dioxygen gas to give $80 \cdot g$ $80*g$ of sulfur trioxide.

Clearly, given this equation, there is excess dioxygen gas. The given molar quantity of sulfur dioxide requires only $1.6 \cdot m o l$ $1.6*mol$ dioxygen gas.

You will have to calculate the mass of product yourself.

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What is the limiting reagent when 3.10 mol of $S O_{2}$ $SO_2$ reacts with 2.75 mol of $O_{2}$ $O_2$ according to the equation: $S O_{2} + O_{2} \to S O_{3}$ $SO_2 + O_2 -> SO_3$ ?

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What is the limiting reagent when 3.10 mol of SO2SO_2 reacts with 2.75 mol of O2O_2 according to the equation: SO2+O2→SO3SO_2 + O_2 -> SO_3?

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What is the limiting reagent when 3.10 mol of $S O_{2}$ $SO_2$ reacts with 2.75 mol of $O_{2}$ $O_2$ according to the equation: $S O_{2} + O_{2} \to S O_{3}$ $SO_2 + O_2 -> SO_3$ ?