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Magnesium metal reacts with nitric acid to produce magnesium nitrate and hydrogen gas according to the following balanced reaction equation:
[tex]\[ \text{Mg} + 2\text{HNO}_3 \rightarrow \text{Mg}\left(\text{NO}_3\right)_2 + \text{H}_2 \][/tex]

Given:
- [tex]\( 0.2 \, \text{dm}^3 \)[/tex] of nitric acid with a mass of [tex]\( 0.2 \, \text{g} \)[/tex] reacts completely with magnesium metal.

1. Calculate the percentage of nitrogen in [tex]\(\text{Mg}\left(\text{NO}_3\right)_2\)[/tex].
2. Calculate the concentration of the nitric acid that reacts.
3. Calculate the mass of magnesium that reacts.
4. Calculate the volume of hydrogen gas produced at STP.
5. Calculate the number of oxygen atoms present in nitric acid.


Sagot :

Certainly! Let's go through each part of the question step by step.

### Step 1: Calculate the Percentage Nitrogen in [tex]\( \text{Mg(NO}_3\text{)}_2 \)[/tex]

To determine the percentage of nitrogen in magnesium nitrate, we need to consider the molar masses involved.

- The molar mass of magnesium nitrate ([tex]\(\text{Mg(NO}_3\text{)}_2\)[/tex]) is approximately 148.32 g/mol.
- Each nitrogen atom has a molar mass of approximately 14.01 g/mol.
- There are 2 nitrogen atoms in magnesium nitrate.

The mass of nitrogen in one mole of [tex]\(\text{Mg(NO}_3\text{)}_2\)[/tex] is [tex]\(2 \times 14.01 = 28.02\)[/tex] g.

To find the percentage of nitrogen, we use the formula:
[tex]\[ \text{Percentage of Nitrogen} = \left(\frac{\text{Mass of Nitrogen in } \text{Mg(NO}_3\text{)}_2}{\text{Molar Mass of } \text{Mg(NO}_3\text{)}_2}\right) \times 100 \][/tex]
[tex]\[ \text{Percentage of Nitrogen} = \left(\frac{28.02}{148.32}\right) \times 100 \approx 18.89\% \][/tex]

So, the percentage of nitrogen in [tex]\(\text{Mg(NO}_3\text{)}_2\)[/tex] is approximately 18.89%.

### Step 2: Calculate the Concentration of the Nitric Acid that Reacts

We are given that the nitric acid has a concentration of 128 mol/L.

### Step 3: Calculate the Mass of Magnesium that Reacts

From the given balanced reaction:
[tex]\[ \text{Mg} + 2\text{HNO}_3 \rightarrow \text{Mg(NO}_3\text{)}_2 + \text{H}_2 \][/tex]

We know 1 mole of magnesium reacts with 2 moles of nitric acid. Given the concentration of nitric acid is 128 mol/L, we can calculate the moles of magnesium that react:
[tex]\[ \text{Moles of Mg} = \frac{128}{2} = 64 \text{ moles} \][/tex]

The molar mass of magnesium ([tex]\(\text{Mg}\)[/tex]) is approximately 24.305 g/mol.

The mass of magnesium reacting can be found by:
[tex]\[ \text{Mass of Mg} = 64 \text{ moles} \times 24.305 \text{ g/mol} = 1555.52 \text{ g} \][/tex]

So, the mass of magnesium that reacts is approximately 1555.52 grams.

### Step 4: Calculate the Volume of Hydrogen Gas Produced at 570K

Given the volume of hydrogen gas produced is 570 dm³ at 570K. To find the volume at standard temperature (273.15K), we employ the ideal gas law properties, specifically using the ratio of volumes to temperatures (as pressure and volume are proportional to temperature):

[tex]\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \][/tex]

Where:
- [tex]\( V_1 = 570 \text{ dm}^3 \)[/tex]
- [tex]\( T_1 = 570 \text{ K} \)[/tex]
- [tex]\( T_2 = 273.15 \text{ K} \)[/tex]

Thus,
[tex]\[ V_2 = V_1 \times \frac{T_2}{T_1} = 570 \text{ dm}^3 \times \frac{273.15 \text{ K}}{570 \text{ K}} \approx 273.15 \text{ dm}^3 \][/tex]

So, the volume of hydrogen gas produced at standard temperature and pressure (STP) is approximately 273.15 dm³.

### Step 5: Calculate the Number of Hydrogen Atoms Present in Nitric Acid

Each molecule of nitric acid ([tex]\(\text{HNO}_3\)[/tex]) contains 1 hydrogen atom.

Given:
- The concentration of nitric acid is 128 mol/L.
- Using Avogadro's number ([tex]\(6.022 \times 10^{23}\)[/tex] atoms per mole).

The total number of hydrogen atoms is:
[tex]\[ \text{Number of hydrogen atoms} = 128 \text{ moles} \times 6.022 \times 10^{23} \text{ atoms/mole} \times 2 \][/tex]
[tex]\[ \text{Number of hydrogen atoms} \approx 1.541632 \times 10^{26} \][/tex]

Thus, the number of hydrogen atoms present in the nitric acid is approximately [tex]\(1.541632 \times 10^{26}\)[/tex].

In summary:
1. The percentage of nitrogen in [tex]\( \text{Mg(NO}_3\text{)}_2 \)[/tex] is approximately 18.89%.
2. The concentration of the nitric acid that reacts is 128 mol/L.
3. The mass of magnesium that reacts is approximately 1555.52 grams.
4. The volume of hydrogen gas produced at 570K is 273.15 dm³ at standard temperature.
5. The number of hydrogen atoms present in nitric acid is approximately [tex]\(1.541632 \times 10^{26}\)[/tex].