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Serggg [28]
3 years ago
14

Iodine, I2, is a solid at room temperature but sublimes (converts from a solid into a gas) when warmed. What is the temperature

(in K) in a 71.4 mL bulb that contains 0.276 g of I2 vapor at a pressure of 0.478 atm?
Chemistry
1 answer:
Stels [109]3 years ago
3 0
<h3>Answer:</h3>

382.63 K

<h3>Explanation:</h3>

We are given;

  • Volume of Iodine as 71.4 mL
  • Mass of Iodine as 0.276 g
  • Pressure of Iodine as 0.478 atm

We are required to calculate the temperature of Iodine

  • We are going to use the ideal gas equation;
  • According to the ideal gas equation; PV = nRT, where R is the ideal gas constant, 0.082057 L.atm/mol.K.
  • Rearranging the formula;

T = PV ÷ nR

But, n, the number of moles = Mass ÷ Molar mass

Molar mass of iodine = 253.8089 g/mol

Thus, n = 0.276 g ÷ 253.8089 g/mol

           = 0.001087 moles

Therefore;

T = (0.478 atm × 0.0714 L) ÷ (0.001087 moles × 0.082057)

  = 382.63 K

Thus, the temperature of Iodine in Kelvin is 382.63 K

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<u>Answer:</u> The limiting reactant is magnesium and mass of excess reactant present in the vessel is 96.35 grams.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For magnesium:</u>

Given mass of magnesium = 41.0 g

Molar mass of magnesium = 24 g/mol

Putting values in equation 1, we get:

\text{Moles of magnesium}=\frac{41.0g}{24g/mol}=1.708mol

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Given mass of iron(III) chloride = 175.0 g

Molar mass of iron(III) chloride = 162.2 g/mol

Putting values in equation 1, we get:

\text{Moles of iron(III) chloride}=\frac{175g}{162.2g/mol}=1.708mol

The chemical equation for the reaction of magnesium and iron(III) chloride follows:

3Mg+2FeCl_3\rightarrow 3MgCl_2+2Fe

By Stoichiometry of the reaction:

3 moles of magnesium reacts with 2 moles of iron(III) chloride

So, 1.708 moles of magnesium will react with = \frac{2}{3}\times 1.708=1.114mol of iron(III) chloride

As, given amount of iron(III) chloride is more than the required amount. So, it is considered as an excess reagent.

Thus, magnesium is considered as a limiting reagent because it limits the formation of product.

Moles of excess reactant left (iron(III) chloride) = [1.708 - 1.114] = 0.594 moles

Now, calculating the mass of iron(III) chloride from equation 1, we get:

Molar mass of iron(III) chloride = 162.2 g/mol

Moles of iron(III) chloride = 0.594 moles

Putting values in equation 1, we get:

0.594mol=\frac{\text{Mass of iron(III) chloride}}{162.2g/mol}\\\\\text{Mass of iron(III) chloride}=(0.594mol\times 162.2g/mol)=96.35g

Hence, the limiting reactant is magnesium and mass of excess reactant present in the vessel is 96.35 grams.

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<h3>Explanation:</h3>
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2I⁻(aq) + Cr²⁺(aq) → CrI₂(s)

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