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3 years ago

Calculate the molarity of 198 g of barium iodide (Bal2) in 2.0 l of solution

Chemistry

1 answer:

nikitadnepr [17]3 years ago

7 0

Answer: The molarity of 198 g of barium iodide $(BaI_{2})$ in 2.0 L of solution is 0.253 M.

Explanation:

Given: Mass = 198 g

Volume = 2.0 L

Molarity is the number of moles of solute present in liter of a solution.

Moles is the mass of a substance divided by its molar mass. So, moles of barium iodide (molar mass = 391.136 g/mol) is calculated as follows.

$Moles = \frac{mass}{molar mass}\\= \frac{198 g}{391.136 g/mol}\\= 0.506 mol$

Now, molarity is calculated as follows.

$Molarity = \frac{moles}{Volume (in L)}\\= \frac{0.506 mol}{2.0 L}\\= 0.253 M$

Thus, we can conclude that the molarity of 198 g of barium iodide $(BaI_{2})$ in 2.0 L of solution is 0.253 M.

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A sample of octane undergoes combustion according to the equation 2 C8H18 + 25 O2 → 16 CO2 + 18 H2O ΔH°rxn = -11018 kJ. What mas

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Answer:

$\large \boxed{\text{528.7 g} }$

Explanation:

It often helps to write the heat as if it were a reactant or a product in the thermochemical equation.

Then you can consider it to be 11018 "moles" of "kJ"

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

M_r: 32.00

2C₈H₁₈ + 25O₂ ⟶ 16CO₂ + 8H₂O + 11 018 kJ

n/mol: 7280

1. Moles of O₂

The molar ratio is 25 mol O₂:11 018 kJ

$\text{Moles of O}_{2} = \text{7280 kJ} \times \dfrac{\text{25 mol O}_{2}}{\text{11 018 kJ}} = \text{16.52 mol O}_{2}$

2. Mass of O₂

$\text{Mass of C$_{8}$H}_{18} = \text{16.52 mol O}_{2} \times \dfrac{\text{32.00 g O}_{2}}{\text{1 mol O}_{2}} = \textbf{528.6 g O}_{2}\\\text{The reaction requires $\large \boxed{\textbf{528.67 g O}_{2}}$}$

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