3 years ago

Help I need to finish my chemistry report.

Chemistry

1 answer:

Alexeev081 [22]3 years ago

4 0

The thing that they have in common is that they are all non metals

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During the chemical reaction given below 21.71 grams of each reagent were allowed to react. Determine how many grams of the exce

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Answer: 16.32 g of $O_2$ as excess reagent are left.

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} SO_2=\frac{21.71g}{64g/mol}=0.34mol$

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According to stoichiometry :

2 moles of $SO_2$ require = 1 mole of $O_2$

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Thus $SO_2$ is the limiting reagent as it limits the formation of product and $O_2$ is the excess reagent.

Moles of $O_2$ left = (0.68-0.17) mol = 0.51 mol

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Thus 16.32 g of $O_2$ as excess reagent are left.

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