How many moles of ethane (C2H6) would be needed to react with 62.3 grams of oxygen gas

Why is it important to add coefficients and not change the subscripts when balancing a chemical equation?

Artemon [7]

Because of making the equestion to be more stable and form the best compound according to it's reactant.

5 0

4 years ago

When nitrogen dioxide (NO2) from car exhaust combines with water in the air, it forms nitric acid (HNO3), which causes acid rain

yKpoI14uk [10]

Answer:

For a: The number of molecules of nitrogen dioxide is $4.52\times 10^{23}$

For b: The mass of nitric acid formed is 54.81 grams

For c: The mass of nitric acid formed is 206 grams

Explanation:

The given chemical reaction follows:

$3NO_2(g)+H_2O(l)\rightarrow 2HNO_3(aq.)+NO(g)$

For a:

By Stoichiometry of the reaction:

1 mole of water reacts with 3 moles of nitrogen dioxide

So, 0.250 moles of water will react with $\frac{3}{1}\times 0.250=0.75mol$ of nitrogen dioxide

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of molecules.

So, 0.75 moles of nitrogen dioxide will contain $0.75\times 6.022\times 10^{23}=4.52\times 10^{23}$ number of molecules

Hence, the number of molecules of nitrogen dioxide is $4.52\times 10^{23}$

For b:

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

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

Given mass of nitrogen dioxide = 60.0 g

Molar mass of nitrogen dioxide = 46 g/mol

Putting values in equation 1, we get:

$\text{Moles of nitrogen dioxide}=\frac{60.0g}{46g/mol}=1.304mol$

By Stoichiometry of the reaction:

3 moles of nitrogen dioxide produces 2 mole of nitric acid

So, 1.304 moles of nitrogen dioxide will produce = $\frac{2}{3}\times 1.304=0.870$ moles of nitric acid

Now, calculating the mass of nitric acid from equation 1, we get:

Molar mass of nitric acid = 63 g/mol

Moles of nitric acid = 0.870 moles

Putting values in equation 1, we get:

$0.870mol=\frac{\text{Mass of nitric acid}}{63g/mol}\\\\\text{Mass of nitric acid}=(0.870mol\times 63g/mol)=54.81g$

Hence, the mass of nitric acid formed is 54.81 grams

For c:

For nitrogen dioxide:

Given mass of nitrogen dioxide = 225 g

Molar mass of nitrogen dioxide = 46 g/mol

Putting values in equation 1, we get:

$\text{Moles of nitrogen dioxide}=\frac{225g}{46g/mol}=4.90mol$

For water:

Given mass of water = 55.2 g

Molar mass of water = 18 g/mol

Putting values in equation 1, we get:

$\text{Moles of water}=\frac{55.2g}{18g/mol}=3.06mol$

By Stoichiometry of the reaction:

3 moles of nitrogen dioxide reacts with 1 mole of water

So, 4.90 moles of nitrogen dioxide will react with = $\frac{1}{3}\times 4.90=1.63mol$ of water

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

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

By Stoichiometry of the reaction:

3 mole of nitrogen dioxide produces 2 moles of nitric acid

So, 4.90 moles of nitrogen dioxide will produce $\frac{2}{3}\times 4.90=3.27mol$ of nitric acid

Now, calculating the mass of nitric acid from equation 1, we get:

Molar mass of nitric acid = 63 g/mol

Moles of nitric acid = 3.27 moles

Putting values in equation 1, we get:

$3.27mol=\frac{\text{Mass of nitric acid}}{63g/mol}\\\\\text{Mass of nitric acid}=(3.27mol\times 63g/mol)=206g$

Hence, the mass of nitric acid formed is 206 grams

7 0

3 years ago

What does the 3 indicate in 1s22s22p63s1?

Dahasolnce [82]

The 3 indicates the third electron shell. (Which has only 1 electron in it in this configuration)

Hope this helps! :)

6 0

3 years ago

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A city continuously disposes of effluent from a wastewater treatment plant into a river. The minimum flow in the river is 130 m3

Vera_Pavlovna [14]

Answer:

$2.54\ \text{mg/L}$

Explanation:

C = Allowable concentration = 1.1 mg/L

$Q_1$ = Flow rate of river = $130\ \text{m}^/\text{s}$

$Q_2$ = Discharge from plant = $37\ \text{m}^3/\text{s}$

$C_1$ = Background concentration = 0.69 mg/L

$C_2$ = Maximum concentration that of the pollutant

The concentration of the mixture will be

$C=\dfrac{Q_1C_1+Q_2C_2}{Q_1+Q_2}\\\Rightarrow C_2=\dfrac{C(Q_1+Q_2)-Q_1C_1}{Q_2}\\\Rightarrow C_2=\dfrac{1.1(130+37)-130\times 0.69}{37}\\\Rightarrow C_2=2.54\ \text{mg/L}$

The maximum concentration that of the pollutant (in mg/L) that can be safely discharged from the wastewater treatment plant is $2.54\ \text{mg/L}$ .

6 0

3 years ago

The periodic table

astra-53 [7]

it's Lithium or Li

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5 0

3 years ago

Read 2 more answers