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Chapter 9: Problem 115

Consider a \(5.00-\mathrm{g}\) sample of silver nitrate, \(\mathrm{AgNO}_{3}(s)\) (a) How many moles of \(\mathrm{AgNO}_{3}\) are in this sample? (b) How many moles of \(\mathrm{O}\) are in this sample? (c) How many grams of \(\mathrm{N}\) are in this sample? (d) How many Ag atoms are in this sample?

Short Answer

Expert verified
a) \(0.0294 \textrm{ mol}\) of AgNO3 b) \(0.0882 \textrm{ mol}\) of oxygen c) \(0.411 \textrm{ g}\) of nitrogen d) \(1.77 \times 10^{22} \textrm{ Ag atoms}\)

Step by step solution

01

Calculate the molar mass of AgNO3

We need to add the molar masses of one Ag atom, one N atom, and three O atoms to find the molar mass of AgNO3. \[M_{AgNO_3} = M_{Ag} + M_{N} + 3 \times M_{O}\] Silver (Ag) has a molar mass of approximately 107.87 g/mol, nitrogen (N) has a molar mass of approximately 14.01 g/mol, and oxygen (O) has a molar mass of approximately 16.00 g/mol. \[M_{AgNO_3} = 107.87 \textrm{ g/mol} + 14.01 \textrm{ g/mol} + 3\times 16.00 \textrm{ g/mol} = 169.88 \textrm{ g/mol}\]
02

Find the number of moles of AgNO3 in the sample

To find the number of moles of AgNO3 in the 5.00 g sample, we will use the following equation: n_AgNO3 = (mass of AgNO3) / (molar mass of AgNO3) \[n_{AgNO_3} = \frac{5.00 \textrm{ g}}{169.88 \textrm{ g/mol}} = 0.0294 \textrm{ mol}\]
03

Determine the number of moles of oxygen and the mass of nitrogen in the sample

Since there are three oxygen atoms in one molecule of AgNO3, the number of moles of oxygen in the sample is: \[n_{O} = 3 \times n_{AgNO_3} = 3 \times 0.0294 \textrm{ mol} = 0.0882 \textrm{ mol}\] Now, to find the mass of nitrogen in the sample, we will use the molar ratio of nitrogen to AgNO3 and the molar mass of nitrogen: \[m_N = n_{AgNO_3} \times M_N = 0.0294 \textrm{ mol} \times 14.01 \textrm{ g/mol} = 0.411 \textrm{ g}\]
04

Calculate the number of Ag atoms in the sample

Using Avogadro's number, we will find the number of Ag atoms in the sample: \[N_{Ag} = n_{AgNO_3} \times N_A = 0.0294 \textrm{ mol} \times (6.022 \times 10^{23} \textrm{ molecules/mol}) = 1.77 \times 10^{22} \textrm{ Ag atoms}\] The answers to the exercise are: a) There are 0.0294 moles of AgNO3 in the sample. b) There are 0.0882 moles of oxygen in the sample. c) There are 0.411 grams of nitrogen in the sample. d) There are 1.77 x 10^22 Ag atoms in the sample.

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Most popular questions from this chapter

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