Chemistry Stoichiometry Question

CH4 with pressure 1 atm and volume 10 liter at 27°C is passed into a reactor with 20% excess oxygen, how many moles of oxygen is

BaLLatris [955]

Answer : The moles of $O_2$ left in the products are 0.16 moles.

Explanation :

First we have to calculate the moles of $CH_4$ .

Using ideal gas equation:

$PV=nRT$

where,

P = pressure of gas = 1 atm

V = volume of gas = 10 L

T = temperature of gas = $27^oC=273+27=300K$

n = number of moles of gas = ?

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

$(1atm)\times (10L)=n\times (0.0821L.atm/mol.K)\times (300K)$

$n=0.406mole$

Now we have to calculate the moles of $O_2$ .

The balanced chemical reaction will be:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

From the balanced reaction we conclude that,

As, 1 mole of $CH_4$ react with 2 moles of $O_2$

So, 0.406 mole of $CH_4$ react with $2\times 0.406=0.812$ moles of $O_2$

Now we have to calculate the excess moles of $O_2$ .

$O_2$ is 20 % excess. That means,

Excess moles of $O_2$ = $\frac{(100 + 20)}{100}$ × Required moles of $O_2$

Excess moles of $O_2$ = 1.2 × Required moles of $O_2$

Excess moles of $O_2$ = 1.2 × 0.812 = 0.97 mole

Now we have to calculate the moles of $O_2$ left in the products.

Moles of $O_2$ left in the products = Excess moles of $O_2$ - Required moles of $O_2$

Moles of $O_2$ left in the products = 0.97 - 0.812 = 0.16 mole

Therefore, the moles of $O_2$ left in the products are 0.16 moles.

7 0

3 years ago

Which gas below has the largest number of moles at STP?

ArbitrLikvidat [17]

Answer:

a.) 22.4 L Ne.

Explanation:

It is known that every 1.0 mol of any gas occupies 22.4 L.

For the options:

22.4 L Ne:

It represents 1.0 mol of Ne.

20 L Ar:

using cross multiplication:

1.0 mol occupies → 22.4 L.

??? mol occupies → 20 L.

The no. of moles of (20 L) Ar = (1.0 mol)(20 L)/(22.4 L) = 0.8929 mol.

2.24 L Xe:

using cross multiplication:

1.0 mol occupies → 22.4 L.

??? mol occupies → 2.24 L.

The no. of moles of (2.24 L) Xe = (1.0 mol)(2.24 L)/(22.4 L) = 0.1 mol.

So, the gas that has the largest number of moles at STP is: a.) 22.4 L Ne.

6 0

3 years ago

How many atoms are in 8.28 moles of aluminum?

Lilit [14]

Answer:

49.86 × 10²³ atoms of Al

Explanation:

Given data:

Number of moles of Al = 8.28 mol

Number of atoms = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

18 g of water = 1 mole = 6.022 × 10²³ molecules of water

1.008 g of hydrogen = 1 mole = 6.022 × 10²³ atoms of hydrogen

For 8.28 moles of Al:

1 mole = 6.022 × 10²³ atoms of Al

8.28 mol×6.022 × 10²³ atoms / 1mol

49.86 × 10²³ atoms of Al

4 0

3 years ago

A student makes a mistake while preparing a vitamin C sample for titration and adds the potassium iodide solution twice. How wil

LiRa [457]

The added KI does not have any impact

The reaction invovles Titration of vitaminc ( Ascorbic acid)

ascorbic acid + I₂ → 2 I⁻ + dehydroascorbic acid

the excess iodine is free reacts with the starch indicator, forming the blue-black starch-iodine complex.

This is the endpoint of the titration. since alreay excess KI is added ( the source of Iodine), it does not have an influence.

Answer B

Hope this helps!

5 0

4 years ago

A pressure of 125,400 Pa is equal to kPa.

dalvyx [7]

Answer:

it is equal to 125 kPa

Explanation:

move the decimal 3 to the left

3 0

3 years ago