Question

Number of moles of O2 required to completely react with 75 grams of Sb?

Answer 1

Answer on the Question #63294, Chemistry / Other

Number of moles of $\mathrm{O}_2$ required to completely reacts with 75 grams of Sb?

Solution:

Oxidation of Sb completely occurs by the following reaction:

$$3 \mathrm{O}_2 + 4 \mathrm{Sb} = 2 \mathrm{Sb}_2\mathrm{O}_3$$

The mole number of O2 is equal to mole number of Sb by this equation (it follows from the reaction):

$$3n(O_2) = 4n(Sb)$$ $$n(O_2) = \frac{4}{3}n(Sb)$$

Let's define the mole number of the Sb:

$$n(Sb) = \frac{m(Sb)}{M(Sb)} = \frac{75g}{\frac{121.8g}{mol}} = 0.62\ \mathrm{mol}$$

Now we can calculate the number of moles of O2 required to completely reacts with 75 grams of Sb:

$$n(O_2) = \frac{4}{3}n(Sb) = \frac{4}{3}0.62\ \mathrm{mol} = 0.83\ \mathrm{mol}$$

Answer: the number of moles of $\mathrm{O}_2$ corresponds to 0.83 moles.

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