If 146.5g of chromium(III) chlorate decomposes what mass of oxygen can be produced?

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If 146.5g of chromium(III) chlorate decomposes what mass of oxygen can be produced?

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Answer 1 · 2018-02-24T03:29:42

The reaction will produce ${69.77 g O}_{2}$ $"69.77 g O"_2$ .

Explanation:

You know that you will need a balanced chemical equation with masses, moles, and molar masses, etc.

Step 1. Assemble all the information in one place

$M_{r} : m m m 302.35 m m m m m m m l l 32.00$ $M_text(r):color(white)(mmm)302.35color(white)(mmmmmmmll)32.00$
$m m m m 2Cr {({ClO}_{3})}_{3} \to {2CrCl}_{3} + {9O}_{2}$ $color(white)(mmmm)"2Cr"("ClO"_3)_3 → "2CrCl"_3 + "9O"_2$
$m /g : m m l l 146.5$ $m"/g": color(white)(mmll)146.5$

Step 2. Calculate the moles of $Cr {({ClO}_{3})}_{3}$ $"Cr"("ClO"_3)_3$

$"Moles of Cr"("ClO"_3)_3 = 146.5 color(red)(cancel(color(black)("g Cr"("ClO"_3)_3))) × ("1 mol Cr"("ClO"_3)_3)/(302.35 color(red)(cancel(color(black)("g Cr"("ClO"_3)_3)))) = 0.4845 color(white)(l) "mol Cr"("ClO"_3)_3$

Step 3 Calculate the moles of $"O"_2$

$"Moles of O"_2 = 0.4845 color(red)(cancel(color(black)("mol Cr"("ClO"_3)_3)))× ("9 mol O"_2)/(2 color(red)(cancel(color(black)("mol Cr"("ClO"_3)_3)))) = "2.180 mol O"_2$

Step 3. Calculate the mass of $"O"_2$

$"Mass of O"_2 = 2.180 color(red)(cancel(color(black)("mol O"_2))) × "32.00 g O"_2/(1 color(red)(cancel(color(black)("mol O"_2)))) = "69.77 g O"_2$

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If 146.5g of chromium(III) chlorate decomposes what mass of oxygen can be produced?

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