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Convert the following to moles:

a) 128.400 mg de Ag

b) [tex]0.74 \, \text{kg}[/tex] de [tex]HNO_3[/tex]

c) 940 g de [tex]FeCl_3[/tex]

Answer :

Sure, let's convert the given amounts to moles step-by-step.

a) 128.400 mg de Ag

1. Convert the mass from milligrams to grams:
[tex]\[
\text{mass of Ag in grams} = \frac{128.400 \, \text{mg}}{1000} = 0.1284 \, \text{g}
\][/tex]

2. Determine the molar mass of Ag (Silver):
[tex]\[
\text{molar mass of Ag} = 107.8682 \, \text{g/mol}
\][/tex]

3. Convert the mass to moles:
[tex]\[
\text{moles of Ag} = \frac{0.1284 \, \text{g}}{107.8682 \, \text{g/mol}} \approx 0.0012 \, \text{moles}
\][/tex]

b) \[tex]$0.74 \, \text{kg} \$[/tex] de \[tex]$ HNO_3 \$[/tex]

1. Convert the mass from kilograms to grams:
[tex]\[
\text{mass of HNO}_3 \text{ in grams} = 0.74 \, \text{kg} \times 1000 = 740 \, \text{g}
\][/tex]

2. Determine the molar mass of HNO₃ (Nitric acid):
[tex]\[
\text{molar mass of HNO}_3 = 1.00784 (\text{H}) + 14.0067 (\text{N}) + 3 \times 15.9994 (\text{O}) = 63.01274 \, \text{g/mol}
\][/tex]

3. Convert the mass to moles:
[tex]\[
\text{moles of HNO}_3 = \frac{740 \, \text{g}}{63.01274 \, \text{g/mol}} \approx 11.7437 \, \text{moles}
\][/tex]

c) 940 g de FeCl₃

1. The mass is already given in grams:
[tex]\[
\text{mass of FeCl}_3 = 940 \, \text{g}
\][/tex]

2. Determine the molar mass of FeCl₃ (Iron(III) chloride):
[tex]\[
\text{molar mass of FeCl}_3 = 55.845 (\text{Fe}) + 3 \times 35.453 (\text{Cl}) = 162.204 \, \text{g/mol}
\][/tex]

3. Convert the mass to moles:
[tex]\[
\text{moles of FeCl}_3 = \frac{940 \, \text{g}}{162.204 \, \text{g/mol}} \approx 5.795 \, \text{moles}
\][/tex]

So, the answers are:
- For 128.400 mg of Ag: approximately 0.0012 moles.
- For \[tex]$0.74 \, \text{kg} \$[/tex] of \[tex]$ HNO_3 \$[/tex]: approximately 11.7437 moles.
- For 940 g of FeCl₃: approximately 5.795 moles.

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