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2 years ago

Which sample contains more atoms: 3.89 g of nickel or 6.61 g of silver? defend your answer

Chemistry

1 answer:

Lubov Fominskaja [6]2 years ago

4 0

Nickel contains more atoms.

<u>Explanation:</u>

Number of atoms is determined by the Avogadro number.

1 mole of substance contains 6.022 X 10²³ atoms.

Given:

Mass of Nickel, Ni = 3.89g

Molecular weight of Ni = 58.6934g

Number of moles of Ni = 3.89/58.6934g

Number of atoms of Ni = $\frac{3.89 X 6.022 X 10^2^3}{58.6934}$

Number of atoms of Ni = 0.39 X 10²³

Mass of silver, Ag = 6.61g

Molecular weight of Ag = 107.8682g

Number of moles of Ag = 6.61/107.8682

Number of atoms of Ag = $\frac{6.61 X 6.022 X 10^2^3}{107.8682}$

Number of atoms of Ag = 0.36 X 10²³

Therefore, Nickel contains for atoms.

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2 years ago

How many liters of hydrogen gas will be produced at STP from the reaction of 7.179×10^23 atoms of magnesium with 54.219g of phos

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Answer: The volume of hydrogen gas produced will be, 12.4 L

Explanation : Given,

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Molar mass of $H_3PO_4$ = 98 g/mol

First we have to calculate the moles of $H_3PO_4$ and $Mg$ .

$\text{Moles of }H_3PO_4=\frac{\text{Given mass }H_3PO_4}{\text{Molar mass }H_3PO_4}$

$\text{Moles of }H_3PO_4=\frac{54.219g}{98g/mol}=0.553mol$

and,

$\text{Moles of }Mg=\frac{7.179\times 10^{23}}{6.022\times 10^{23}}=1.19mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$3Mg+2H_3PO_4\rightarrow Mg(PO_4)_2+3H_2$

From the balanced reaction we conclude that

As, 3 mole of $Mg$ react with 2 mole of $H_3PO_4$

So, 0.553 moles of $Mg$ react with $\frac{2}{3}\times 0.553=0.369$ moles of $H_3PO_4$

From this we conclude that, $H_3PO_4$ is an excess reagent because the given moles are greater than the required moles and $Mg$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $H_2$

From the reaction, we conclude that

As, 3 mole of $Mg$ react to give 3 mole of $H_2$

So, 0.553 mole of $Mg$ react to give 0.553 mole of $H_2$

Now we have to calculate the volume of $H_2$ gas at STP.

As we know that, 1 mole of substance occupies 22.4 L volume of gas.

As, 1 mole of hydrogen gas occupies 22.4 L volume of hydrogen gas

So, 0.553 mole of hydrogen gas occupies $0.553\times 22.4=12.4L$ volume of hydrogen gas

Therefore, the volume of hydrogen gas produced will be, 12.4 L

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3 years ago