Question #64391

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Question #64391

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Answer 1 · 2018-01-19T09:41:44

$4 moles$ $"4 moles"$ .

Explanation:

Let's start with the skeleton equation:

$K C l O_{3} \to K C l + O_{2}$ $KClO_3 -> KCl + O_2$

Then, balance both sides:

$2 K C l O_{3} \to 2 K C l + 3 O_{2}$ $2KClO_3 -> 2KCl + 3O_2$

From this, we know that $2$ $2$ moles of $K C l O_{3}$ $KClO_3$ corresponds to $3$ $3$ moles of $O_{2}$ $O_2$ . That means we just need to find out the number of moles of $O_{2}$ $O_2$ in $200g$ $"200g"$ to see how many moles of $K C l O_{3}$ $KClO_3$ decomposed!

The mass of one mole, or the molar mass, of $O_{2}$ $O_2$ is $16.00 + 16.00 = 32.00g$ $16.00+16.00="32.00g"$ , because the mass of one mole of $O$ $O$ is $16.00g$ $"16.00g"$ .
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So, $200g$ $"200g"$ of $O_{2}$ $O_2$ is:
$\frac{200}{32.00} = 6.25 moles$ $200/32.00="6.25 moles"$

Again, $3$ $3$ moles of $O_{2}$ $O_2$ corresponds to $2$ $2$ moles of $K C l O_{3}$ $KClO_3$ .
Therefore, $6.25 moles$ $"6.25 moles"$ of $O_{2}$ $O_2$ corresponds to $2 \cdot \frac{6.25}{3} = 4.16666667 moles$ $2*6.25/3="4.16666667 moles"$ of $K C l O_{3}$ $KClO_3$ .

Finally, because the question had $1$ $1$ significant figure, our answer is $4 moles$ $"4 moles"$ .

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Question #64391

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