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alexira [117]
3 years ago
8

The hormone thyroxine is secreted by the thyroid gland and has the formula: C15H17NO4I4. How many milligrams of Iodine can be ex

tracted from 15.0 grams of thryoxine?
Chemistry
2 answers:
yan [13]3 years ago
7 0

Answer : The mass of iodine extracted can be 9796.7 mg

Explanation : Given,

Mass of thryoxine = 15.0 g

Molar mass of thryoxine = 776.86 g/mole

The molecular formula of thryoxine is, C_{15}H_{11}NO_4I_4

In C_{15}H_{11}NO_4I_4 compound, there are 15 moles of carbon, 11 moles of hydrogen, 1 mole of nitrogen, 4 moles of oxygen and 4 moles of iodine.

First we have to determine the moles of thryoxine.

\text{Moles of thryoxine}=\frac{\text{Mass of thryoxine}}{\text{Molar mass of thryoxine}}=\frac{15.0g}{776.86g/mole}=0.0193moles

Now we have to determine the moles of iodine.

As, 1 mole of thryoxine has 4 moles of iodine

So, 0.0193 mole of thryoxine has 4\times 0.0193=0.0772 moles of iodine

Now we have to determine the mass of iodine.

\text{Mass of iodine}=\text{Moles of iodine}\times \text{Molar mass of iodine}

\text{Mass of iodine}=(0.0772mole)\times (126.9g/mole)=9.7967g=9796.7mg

conversion used : (1 g = 1000 mg)

Therefore, the mass of iodine extracted can be 9796.7 mg

Ede4ka [16]3 years ago
6 0
I'm pretty sure it's 9726 milligrams of iodine. Hope this helps.
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Read 2 more answers
9. Using the balanced equation from Question #8, how many grams of lead will be produced if 2.54 grams of PbS is burned with 1.8
MissTica

Answer: 2.24 grams of Pb

Explanation:

<u>Step 1</u>

Balanced chemical reaction;

2PbS + 3O2 → 2Pb + 2SO3

<u>Step 2</u>

Moles of both PbS and O2

Moles = mass / molar mass

Moles of PbS = 2.54 g / 239.3 g/mol = 0.0108 moles

Moles of O2 = 1.88 / 32 g/mol = 0.0588 moles

<u>Step 3</u>

Finding the limiting reactant.

Limiting reactant, is that reactant which is completely used in the reaction;

If we assume that PbS is the limiting reactant;

We have 0.0588 moles of O2. This needs ( 0.0588 * 2) / 3 = 0.0392 moles of PbS to fully react. But we have only 0.0108 moles of PbS available. That means that the PbS will be completely consumed hence the limiting reactant

If we assume O2 is the limiting reactant;

We have 0.0108 moles of PbS. That needs ( 0.0108 * 3) / 2 = 0.0162 moles of O2. But we have 0.0588 moles of O2 which is in excess further confirming that PbS is the limiting reactant since it will be depleted in the reaction.

<u>Step 4</u>

Moles of lead

For this step we apply the mole ratios with the limiting reactant;

Mole ratio of PbS : Pb = 2 : 2 = 1 : 1

Therefore;

Moles of Pb = (0.0108 moles  * 1 ) 1

Moles of Pb =0.0108 moles

<u>Step 5</u>

Mass of Pb

Mass = moles * molar mass

Mass of Pb =0.0108 moles * 207.2 g/mol

Mass of Pb = 2.24 grams

5 0
2 years ago
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