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Gemiola [76]
2 years ago
6

A mixture of hydrogen (2.02 g) and chlorine (35.90 g) in a container at 300 K has a total gas pressure of 748 mm Hg. What is the

partial atmospheric pressure (atm) of hydrogen in the mixture?
Chemistry
1 answer:
Llana [10]2 years ago
3 0

The partial atmospheric pressure (atm) of hydrogen in the mixture is 0.59 atm.

<h3>How do we calculate the partial pressure of gas?</h3>

Partial pressure of particular gas will be calculated as:

p = nP, where

  • P = total pressure = 748 mmHg
  • n is the mole fraction which can be calculated as:
  • n = moles of gas / total moles of gas

Moles will be calculated as:

  • n = W/M, where
  • W = given mass
  • M = molar mass

Moles of Hydrogen gas = 2.02g / 2.014g/mol = 1 mole

Moles of Chlorine gas = 35.90g / 70.9g/mol = 0.5 mole

Mole fraction of hydrogen = 1 / (1+0.5) = 0.6

Partial pressure of hydrogen = (0.6)(748) = 448.8 mmHg = 0.59 atm

Hence, required partial atmospheric pressure of hydrogen is 0.59 atm.

To know more about partial pressure, visit the below link:
brainly.com/question/15302032

#SPJ1

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Answer:

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Explanation:

Hello,

In this case, for the given chemical reaction, we first identify the limiting reactant by noticing that due to the 1:1 mole ratio for magnesium to iodine the reacting moles must the same, nevertheless, there are only 2.68 moles of magnesium versus 3.56 moles of iodine, for that reason, magnesium is the limiting reactant, so the theoretical turns out:

n_{MgI_2}^{theoretical}=2.68molMg*\frac{1molMgI_2}{1molMg} =2.68molMgI_2

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