# Question a0976

Dec 26, 2017

$= 75 g S {O}_{3}$

#### Explanation:

1. Write and balance the equation
$2 S + 3 {O}_{2} \to 2 S {O}_{3}$
2. Find the molar masses of the involved substances. These values are obtainable from the periodic table; i.e.,
$S = \frac{32 g}{m o l}$
${O}_{2} = \frac{32 g}{m o l}$
$S {O}_{3} = \frac{80 g}{m o l}$
3. Given the masses of the reactants, individual value in moles can be calculated; thus,
$\eta S = 40 \cancel{g S} \times \frac{1 m o l S}{32 \cancel{g S}} = 1.2500 m o l S$
$\eta {O}_{2} = 45 \cancel{g {O}_{2}} \times \frac{1 m o l {O}_{2}}{32 \cancel{g {O}_{2}}} = 1.4063 m o l {O}_{2}$
4. Now, identify which of the reactants limit the production of $S {O}_{3}$ by determining how much each reactant is required when paired with each other. This case, refer to the balanced equation for the mole ratio or to the problem where ratio of the reaction is given as ${2}_{S} : {3}_{{O}_{2}}$; i.e.,
$a . \eta S = 1.2500 m o l$
$= 1.2500 \cancel{m o l S} \times \frac{3 m o l {O}_{2}}{2 \cancel{m o l S}}$
$= 1.875 m o l {O}_{2}$
$\therefore$
$\frac{\eta {O}_{2} \text{ available")/(=1.4063 molO_2)<(etaO_2 " required}}{= 1.875 m o l {O}_{2}}$
color(red)("This means that " 1.2500molS-=1.875molO_2 " for complete reaction. This case, " O_2 " is the limiting reactant."
$b . \eta {O}_{2} = 1.4063 m o l$
$= 1.4063 \cancel{m o l {O}_{2}} \times \frac{2 m o S}{3 \cancel{m o l {O}_{2}}}$
$= 0.9375 m o l S$
$\therefore$
$\frac{\eta S \text{ available")/(=1.2500molS)>(etaS " required}}{= 0.9375 m o l S}$
color(blue)("This means that " 1.4063molO_2-=0.9375molS " for complete reaction. This case, " S " is the x's reactant."#
5. Knowing the limiting reactant will provide the maximum production for $S {O}_{3}$; leaving an x's for the other reactant. With reference to the balanced equation for the mole ratio and the molar mass of $S {O}_{3}$, the theoretical yield for this reaction can be calculated as follows:
$= 1.4063 \cancel{m o l {O}_{2}} \times \frac{2 \cancel{m o l S {O}_{3}}}{3 \cancel{m o l {O}_{2}}} \times \frac{80 g S {O}_{3}}{1 \cancel{m o l S {O}_{3}}}$
$= 75 g S {O}_{3}$