Liquid molecules have a:

How many moles of Cl2 are needed in order to produce 100.0 grams of FeCl3 given the following

dimulka [17.4K]

Answer:

0.92moles

Explanation:

Given reaction:

2Fe + 3Cl₂ → 2FeCl₃

Mass of FeCl₃ = 100g

Unknown:

Number of moles of Cl₂ needed = ?

Solution:

To solve this problem, we work from the known specie to the unknown.

From the mass of the FeCl₃ given, we can solve for the number of moles of the unknown.

Number of moles of FeCl₃;

Number of moles = $\frac{mass}{molar mass}$

Molar mass of FeCl₃ = 56 + 3(35.5) = 162.5g/mol

Number of moles = $\frac{100}{162.5}$ = 0.62mole

From the reaction expression;

2 mole of FeCl₃ is produced from 3 moles of Cl₂

0.62 mole of FeCl₃ will be produced from $\frac{0.62 x 3}{2}$ = 0.92mole

The number of moles of Cl₂ = 0.92moles

5 0

3 years ago

Determine the limiting reactant when 30.0 g of propane, C3H8, is burned with 75.0 g of oxygen.

AnnZ [28]

Answer: The limiting reagent is oxygen gas.

Explanation:

Limiting reagent is defined as the reactant that is present in less amount and it limits the formation of products.

Excess reagent is defined as the reactant which is present in large amount.

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

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

For propane:

Given mass of propane = 30.0 g

Molar mass of propane = 44.1 g/mol

Putting values in equation 1, we get:

$\text{Moles of propane}=\frac{30.0g}{44.1g/mol}=0.680mol$

For oxygen:

Given mass of oxygen = 75.0 g

Molar mass of oxygen = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of propane}=\frac{75.0g}{32g/mol}=2.34mol$

The chemical equation for the combustion of propane follows:

$C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(l)$

By Stoichiometry of the reaction:

5 moles of oxygen gas reacts with 1 mole of propane.

So, 2.34 moles of oxygen gas will react with = $\frac{1}{5}\times 2.34=0.468mol$ of propane

As, given amount of propane is more than the required amount. So, it is considered as an excess reagent.

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

Hence, the limiting reagent is oxygen gas.

3 0

3 years ago

Answer the following questions about the fermentation of glucose (C6H12O6, molar mass 180.2 g/mol)

Yuri [45]

The amount of energy in kilocalories released from 49 g of glucose given the data is -4.4 Kcal

How to determine the mole of glucose

Mass of glucose = 49 g

Molar mass of glucose = 180.2 g/mol

Mole of glucose = ?

Mole = mass / molar mass

Mole of glucose = 49 / 180.2

Mole of glucose = 0.272 mole

How to determine the energy released

C₆H₁₂O₆ →2C₂H₆O + 2CO₂ ΔH = -16 kcal/mol

From the balanced equation above,

1 mole of glucose released -16 kcal of energy

Therefore,

0.272 mole of glucose will release = 0.272 × -16 = -4.4 Kcal

Thus, -4.4 Kcal were released from the reaction

Learn more about stoichiometry:

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6 0

2 years ago

A 650.0 mL solution contains 125 grams of glucose (C6H12O6). If the molar mass of C6H12O6 is 180.16 g/mol, what is the molarity

Aleks [24]

Molarity of a solution is the molar concentration, which is the number of moles of solute in 1 L of solution.

the mass of glucose is - 125 g

number of moles of glucose - 125 g / 180.16 g/mol = 0.694 mol

the number of moles of glucose in 650.0 x 10⁻³ L - 0.694 mol

number of moles of glucose in 1 L - 0.694 mol / 650.0 x 10⁻³ L = 1.068 mol/L

molarity of glucose solution is 1.068 M

4 0

3 years ago

Read 2 more answers

The chemical equation below summarizes cellular respiration. 6O2 + C6H12O6 à 6CO2 + 6H2O + energy The energy that is released by

Kay [80]

Chemical energy is released from the breaking down of glucose.

4 0

3 years ago

Read 2 more answers