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A 3.8-mol sample of [tex][tex]$KClO_3$[/tex][/tex] was decomposed according to the equation below. How many moles of [tex][tex]$O_2$[/tex][/tex] are formed assuming [tex]100 \%[/tex] yield?

[tex]\[ 2 KClO_3(s) \rightarrow 2 KCl(s) + 3 O_2(g) \][/tex]

A. 2.5 mol
B. 3.2 mol
C. 3.8 mol
D. 1.9 mol
E. 5.7 mol

Sagot :

Absolutely! Let's solve the problem step-by-step.

Given the decomposition reaction of potassium chlorate ([tex]\(KClO_3\)[/tex]):

[tex]\[ 2 KClO_3(s) \rightarrow 2 KCl(s) + 3 O_2(g) \][/tex]

To find out how many moles of [tex]\(O_2\)[/tex] are formed from a 3.8-mol sample of [tex]\(KClO_3\)[/tex], we need to analyze the molar relationship from the balanced chemical equation.

1. Identify the molar ratio: According to the balanced chemical equation, 2 moles of [tex]\(KClO_3\)[/tex] decompose to produce 3 moles of [tex]\(O_2\)[/tex].

[tex]\[ 2 \text{ moles } KClO_3 \rightarrow 3 \text{ moles } O_2 \][/tex]

2. Calculate the molar ratio: For 1 mole of [tex]\(KClO_3\)[/tex]:

[tex]\[ \text{Moles of } O_2 = 1 \times \frac{3}{2} = 1.5 \text{ moles of } O_2 \][/tex]

3. Apply the molar ratio to the given sample: Now, we need to determine how many moles of [tex]\(O_2\)[/tex] are produced from a 3.8-mol sample of [tex]\(KClO_3\)[/tex]:

[tex]\[ \text{Moles of } O_2 = 3.8 \text{ moles of } KClO_3 \times \frac{3}{2} = 3.8 \times 1.5 \][/tex]

4. Perform the multiplication:

[tex]\[ 3.8 \times 1.5 = 5.7 \text{ moles of } O_2 \][/tex]

Hence, the number of moles of [tex]\(O_2\)[/tex] formed from a 3.8-mol sample of [tex]\(KClO_3\)[/tex] is [tex]\(5.7\)[/tex] moles.

Therefore, the correct answer is:

[tex]\[ \boxed{5.7 \text{ mol}} \][/tex]