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3 years ago

Consider the titration of 30.0 mL of 0.050 M NH3 with 0.025 M HCl.

Chemistry

1 answer:

Alla [95]3 years ago

6 0

Answer:

60.0mL

Explanation:

For the process of titration , the formula we use is -

M₁V₁ = M₂V₂

Where ,

M₁ = initial concentration

V₁ = initial volume

M₂ = final concentration

V₂ = final volume .

Hence , from the question ,

The given data is ,

M₁ = 0.050 M

V₁ = 30.0mL

M₂ = 0.025 M

V₂ = ?

Now, to determine the unknown quantity , the formula can be applied ,

Hence ,

M₁V₁ = M₂V₂

Putting the respective values ,

0.050 M * 30.0mL = 0.025 M * ?

solving the above equation ,

V₂ = ? = 60.0mL

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A flexible container at an initial volume of 6.13 L contains 2.51 mol of gas. More gas is then added to the container until it r

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Answer:

2.12 moles of gas were added.

Explanation:

We can solve this problem by using Avogadro's law, which states that at constant temperature and pressure:

V₁n₂=V₂n₁

Where in this case:

V₁ = 6.13 L

n₂ = ?

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n₁ = 2.51 mol

We input the data:

6.13 L * n₂ = 11.3 L * 2.51 mol

n₂ = 4.63

As 4.63 moles is the final number of moles, the number of moles added is:

4.63 - 2.51 = 2.12 moles

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What mass, in grams, of CO2 and H2O is formed from 2.55 mol of propane?

oksian1 [2.3K]

Answer:

336.6 grams of CO₂ and 183.6 grams of H₂O are formed from 2.55 moles of propane.

Explanation:

In this case, the balanced reaction is:

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reactant and product participate in the reaction:

C₃H₈: 1 mole
O₂: 5 moles
CO₂: 3 moles
H₂O: 4 moles

Being the molar mass of each compound:

C₃H₈: 44 g/mole
O₂: 16 g/mole
CO₂: 44 g/mole
H₂O: 18 g/mole

Then, by stoichiometry, the following quantities of mass participate in the reaction:

C₃H₈: 1 mole* 44 g/mole= 44 grams
O₂: 5 moles* 16 g/mole= 80 grams
CO₂: 3 moles* 44 g/mole= 132 grams
H₂O: 4 moles* 18 g/mole= 72 grams

So you can apply the following rules of three:

If by stoichiometry 1 mole of C₃H₈ forms 132 grams of CO₂, 2.55 moles of C₃H₈ how much mass of CO₂ will it form?

$mass of CO_{2} =\frac{2.55 moles of C_{3} H_{8}*132 gramsof CO_{2} }{ 1 mole of C_{3} H_{8}}$

mass of CO₂= 336.6 grams

If by stoichiometry 1 mole of C₃H₈ forms 72 grams of H₂O, 2.55 moles of C₃H₈ how much mass of H₂O will it form?

$mass of H_{2}O =\frac{2.55 moles of C_{3} H_{8}*72 gramsof H_{2}O }{ 1 mole of C_{3} H_{8}}$

mass of H₂O= 183.6 grams

336.6 grams of CO₂ and 183.6 grams of H₂O are formed from 2.55 moles of propane.

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3 years ago