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kenny6666 [7]
3 years ago
11

A 0.77 mg sample of nitrogen reacts with chlorine to form 6.615 mg of the chloride. part a what is the empirical formula of the

nitrogen chloride?
Chemistry
2 answers:
AysviL [449]3 years ago
6 0

From the given:

Reacted mass of nitrogen = 0.77 mg

Mass of chlorine = 6.615 mg

Reacted mass of chlorine = 6.615 mg - 0.77 = 5.84 mg

Mass of Nitrogen = \frac{0.77}{14} = 0.055

Mass of chlorine = \frac{5.84}{35.5} = 0.165

Each value is divided by small number

\frac{0.055}{0.055} =1   ; \frac{0.165}{0.055} = 3

N_{1} Cl_{3}

Therefore, empirical formula is NCl_{3}

Shkiper50 [21]3 years ago
4 0

Answer:

The empirical formula of the nitrogen chloride is 2NCl_{3}

Explanation:

Step 1: Form the formation equation of nitrogen chloride. The N and Cl are diatomic molecules, so they have subindices 2. We balance the chemical equation remaining:

N_{2} + 3Cl_{2} ->  2NCl_{3}

Step 2: We calculate the number of moles that are used from N_{2} and the number of moles that we get from  2NCl_{3}.

moles of N_{2} = 0,77mg N_{2} * 1g/(1000mg) * 1/(28 g N_{2}/mol N_{2})

moles of N_{2} = 2,75 * 10^(-5)

moles of  2NCl_{3} = 6,615mg 2NCl3 * 1g/(1000mg) * 1/(28 g  2NCl_{3}/mol  2NCl_{3})

moles of  2NCl_{3} = 5,5*10^(-5)

Step 3: We verify that the number of moles formed is correct using the reaction balance coefficients, taking into account that 1 mol of N_{2} produces 2 moles of  2NCl_{3}.

2.75 * 10^(-5) moles of N_{2} = (5.5*10^(-5) moles of  2NCl_{3}) / 2

2,75 * 10^(-5) moles = 2,75 * 10^(-5) moles

Have a nice day!

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