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nekit [7.7K]
3 years ago
10

Show your work and include answer with units. How many atoms in 55 grams of magnesium

Chemistry
1 answer:
Georgia [21]3 years ago
8 0

Answer:

1.38 x 10^{24} atoms

Explanation:

To get grams to atoms/molecules/particles we need to first convert grams > moles > atoms. This also applies in the other direction, where you have moles as that stepping stone.

Some key information we need:

Magnesium's molar mass= 24g/mol

Avogadro's Number= 6.022 x 10^{23} atoms

Where there is 1 mol per 6.022 x 10^{23}  atoms

*This maybe be something you want to memorize.

Grams to moles using molar mass , then to atoms using avogadro's number.

55g Mg x \frac{1 mol}{24g/mol} x \frac{6.022 x 10^{23} atoms }{1 mol} = 1.38 x 10^{24} atoms

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A reaction of 41.9 g of Na and 30.3 g of Br2 yields 36.4 g of NaBr . What is the percent yield?
Licemer1 [7]

Answer: The percent yield is, 93.4%

Explanation:

First we have to calculate the moles of Na.

\text{Moles of Na}=\frac{\text{Mass of Na}}{\text{Molar mass of Na}}=\frac{41.9g}{23g/mole}=1.82moles

Now we have to calculate the moles of Br_2

{\text{Moles of}Br_2} = \frac{\text{Mass of }Br_2 }{\text{Molar mass of} Br_2} =\frac{30.3g}{160g/mole}=0.189moles

{\text{Moles of } NaBr} = \frac{\text{Mass of } NaBr }{\text{Molar mass of } NaBr} =\frac{36.4g}{103g/mole}=0.353moles

The balanced chemical reaction is,

2Na(s)+Br_2(g)\rightarrow 2NaBr

As, 1 mole of bromine react with = 2 moles of Sodium

So, 0.189 moles of bromine react with = \frac{2}{1}\times 0.189=0.378 moles of Sodium

Thus bromine is the limiting reagent as it limits the formation of product and Na is the excess reagent.

As, 1 mole of bromine give = 2 moles of Sodium bromide

So, 0.189 moles of bromine give = \frac{2}{1}\times 0.189=0.378 moles of Sodium bromide

Now we have to calculate the percent yield of reaction

\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100=\frac{0.353 mol}{0.378}\times 100=93.4\%

Therefore, the percent yield is, 93.4%

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