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3 years ago

Question 4 (1 point) If you have exactly one million atoms how many moles do you have?

Chemistry

1 answer:

rjkz [21]3 years ago

7 0

Answer:

1.66 × 10⁻¹⁸ Moles

Explanation:

As we know one mole of any substance contains 6.022 × 10²³ particles (atoms, ions, molecules or formula units). This number is also called as Avogadro's Number.

The relation between Moles, Number of Atoms and Avogadro's Number is given as,

Number of Moles = Number of Atoms ÷ 6.022 × 10²³ Atoms/mol

Putting values,

Number of Moles = 1.0 × 10⁶ Atoms ÷ 6.022 × 10²³ Atoms/mol

Number of Moles = 1.66 × 10⁻¹⁸ Moles

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A student reacts 5.0 g of sodium with 10.0 g of chlorine and collect 5.24 g of sodium chloride. What is the percent yield of thi

Ede4ka [16]

Answer: The percent yield of this combination reaction is 41.3 %

Explanation : Given,

Mass of $Na$ = 5.0 g

Mass of $Cl_2$ = 10.0 g

Molar mass of $Na$ = 23 g/mol

Molar mass of $Cl_2$ = 71 g/mol

First we have to calculate the moles of $Na$ and $Cl_2$ .

$\text{Moles of }Na=\frac{\text{Given mass }Na}{\text{Molar mass }Na}$

$\text{Moles of }Na=\frac{5.0g}{23g/mol}=0.217mol$

and,

$\text{Moles of }Cl_2=\frac{\text{Given mass }Cl_2}{\text{Molar mass }Cl_2}$

$\text{Moles of }Cl_2=\frac{10.0g}{71g/mol}=0.141mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation will be:

$2Na+Cl_2\rightarrow 2NaCl$

From the balanced reaction we conclude that

As, 2 mole of $Na$ react with 1 mole of $Cl_2$

So, 0.217 moles of $Na$ react with $\frac{0.217}{2}=0.108$ moles of $Cl_2$

From this we conclude that, $Cl_2$ is an excess reagent because the given moles are greater than the required moles and $Na$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NaCl$

From the reaction, we conclude that

As, 2 mole of $Na$ react to give 2 mole of $NaCl$

So, 0.217 mole of $HCl$ react to give 0.217 mole of $NaCl$

Now we have to calculate the mass of $NaCl$

$\text{ Mass of }NaCl=\text{ Moles of }NaCl\times \text{ Molar mass of }NaCl$

Molar mass of $NaCl$ = 58.5 g/mole

$\text{ Mass of }NaCl=(0.217moles)\times (58.5g/mole)=12.7g$

Now we have to calculate the percent yield of this reaction.

Percent yield = $\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100$

Actual yield = 5.24 g

Theoretical yield = 12.7 g

Percent yield = $\frac{5.24g}{12.7g}\times 100$

Percent yield = 41.3 %

Therefore, the percent yield of this combination reaction is 41.3 %

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