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3 years ago

If you have 1.1 moles of magnesium nitrate then how many grams is that

Chemistry

1 answer:

marysya [2.9K]3 years ago

3 0

Answer: 162.8 grams

Explanation:

Magnesium nitrate has a chemical formula of Mg(NO3)2.

Given that:

Number of moles of Mg(NO3)2 = 1.1 moles

Mass in grams of Mg(NO3)2 = ?

For Molar mass of Mg(NO3)2, use atomic mass of magnesium = 24g, nitrogen = 14g, oxygen = 16g

Mg(NO3)2 = 24g + (14g + 16gx3) x 2

= 24g + (14g + 48g) x 2

= 24g + (62g) x 2

= 24g + 124g

= 148g/mol

Now, apply the formula:

Number of moles = Mass in grams / molar mass

1.1 moles = Mass / 148g/mol

Mass = 1.1 moles x 148g/mol

Mass = 162.8 grams

Thus, there are 162.8 grams of magnesium nitrate.

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shusha [124]

214.0560
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3 years ago

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 $Br_2$

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

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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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3 years ago

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Answer:

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See the explanation below, please.

Explanation:

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