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50.0 mL of a 1.25 M solution of nitric acid is added to 1.12 g of solid iron(III) hydroxide, what is the concentration (in M) of iron(III) ions in the final solution?Fe(OH)3(s)  +  3 HNO3(aq)  →  Fe(NO3)3(aq)  +  3 H2O(l)Assume the volume of the solution remains constant. Do not include the units in your answer.  Your answer should have three significant figures.

Question

50.0 mL of a 1.25 M solution of nitric acid is added to 1.12 g of solid iron(III) hydroxide, what is the concentration (in M) of iron(III) ions in the final solution?

Fe(OH)3(s)  +  3 HNO3(aq)  →  Fe(NO3)3(aq)  +  3 H2O(l)

Assume the volume of the solution remains constant. Do not include the units in your answer.  Your answer should have three significant figures.

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Step 1

50.0 mL of a 1.25 M solution of nitric acid is added to 1.12 g of solid iron(III) hydroxide, the concentration (in M) of iron(III) ions in the final solution is to be calculated assuming the volume of the solution remains constant.

Fe(OH), (s) 3 HNO, (aq) > Fe(NO) (aq) + 3 H,O()
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Fe(OH), (s) 3 HNO, (aq) > Fe(NO) (aq) + 3 H,O()

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Step 2

Calculate the moles of Iron (III) oxide, Fe(OH)3, in 1.12 g of solid iron(III) hydroxide -

Given: Mass of Fe(OH)3 = 1.12 g

Molar mass of Fe(OH)3 = 159.69 g/mol

Mass of Fe(OH)
Moles of Fe(OFH); Molar mass of Fe(OH)
1.12 g
Moles of Fe(OH); 159.69 g/mol
. Moles of Fe(OH) = 7.01 x10 moles
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Mass of Fe(OH) Moles of Fe(OFH); Molar mass of Fe(OH) 1.12 g Moles of Fe(OH); 159.69 g/mol . Moles of Fe(OH) = 7.01 x10 moles

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Step 3

Calculate the moles of nitric acid-

Molarity of HNO3 = 1.25 M

Volume ...

Moles of HNO
Molarity of HNO
Volume of solution in L
Moles of HNO
50x10 L
:. Moles of HNO3 = (1.25 M)x (50 x10
1.25 M
L) = 62.5x10
moles
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Moles of HNO Molarity of HNO Volume of solution in L Moles of HNO 50x10 L :. Moles of HNO3 = (1.25 M)x (50 x10 1.25 M L) = 62.5x10 moles

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