How did they get the limiting reactant as O2???

Answer my question please!

Elza [17]

I think it is B! I am so sorry if it’s wrong, love.

5 0

3 years ago

Which of these elements have the same number of valance electrons

solmaris [256]

Both carbon and lead belong to Group IV elements, and thus they have the same number of valence electrons.
<span>In each of the other options, the two elements belong to different groups, and thus they do NOT have the same number of valence electrons.

I hope this helped you, please tell me if I am correct or not <3
</span>

6 0

4 years ago

How many moles of a gas sample are in a 20.0 L container at 373 K and 203 kPa? The gas constant is 8.31 L−kPa/mol−K. A)0.33 mole

12345 [234]

Answer:

Option (C) 1.30 moles

Explanation:

The following data were obtained from the question:

Volume (V) = 20L

Temperature (T) = 373K

Pressure (P) = 203 kPa

Gas constant (R) = 8.31 L.kPa/mol.K.

Number of mole (n) =...?

The number of mole of the gas in the container can obtained by applying the ideal gas equation as illustrated below:

PV = nRT

Divide both side by RT

n = PV /RT

n = 203 x 20 / 8.31 x 373

n = 1.30 mole.

Therefore, 1.30 mole of the gas is present in the container.

4 0

3 years ago

Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . If of water is pro

Lubov Fominskaja [6]

This is an incomplete question, here is a complete question.

Aqueous sulfuric acid (H₂SO₄) reacts with solid sodium hydroxide (NaOH) to produce aqueous sodium sulfate (Na₂SO₄) and liquid water (H₂O) . If 12.5 g of water is produced from the reaction of 72.6 g of sulfuric acid and 77.0 g of sodium hydroxide, calculate the percent yield of water. Round your answer in significant figures.

Answer: The percent yield of water is, 46.8 %

Explanation : Given,

Mass of $H_2SO_4$ = 72.6 g

Mass of $NaOH$ = 77.0 g

Molar mass of $H_2SO_4$ = 98 g/mol

Molar mass of $NaOH$ = 40 g/mol

First we have to calculate the moles of $H_2SO_4$ and $NaOH$ .

$\text{Moles of }H_2SO_4=\frac{\text{Given mass }H_2SO_4}{\text{Molar mass }H_2SO_4}$

$\text{Moles of }H_2SO_4=\frac{72.6g}{98g/mol}=0.741mol$

and,

$\text{Moles of }NaOH=\frac{\text{Given mass }NaOH}{\text{Molar mass }NaOH}$

$\text{Moles of }NaOH=\frac{77.0g}{40g/mol}=1.925mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O$

From the balanced reaction we conclude that

As, 1 mole of $H_2SO_4$ react with 2 mole of $NaOH$

So, 0.741 moles of $H_2SO_4$ react with $0.741\times 2=1.482$ moles of $NaOH$

From this we conclude that, $NaOH$ is an excess reagent because the given moles are greater than the required moles and $H_2SO_4$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $H_2O$

From the reaction, we conclude that

As, 1 mole of $H_2SO_4$ react to give 2 mole of $H_2O$

So, 0.741 moles of $H_2SO_4$ react to give $0.741\times 2=1.482$ moles of $H_2O$

Now we have to calculate the mass of $H_2O$

$\text{ Mass of }H_2O=\text{ Moles of }H_2O\times \text{ Molar mass of }H_2O$

Molar mass of $H_2O$ = 18 g/mole

$\text{ Mass of }H_2O=(1.482moles)\times (18g/mole)=26.68g$

Now we have to calculate the percent yield of water.

$\text{Percent yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100$

Now put all the given values in this formula, we get:

$\text{Percent yield}=\frac{12.5g}{26.68g}\times 100=46.8\%$

Thus, the percent yield of water is, 46.8 %

3 0

3 years ago

How cogeneration reduces fossil fuel?

stepan [7]

Answer:

Along with the saving of fossil fuels, cogeneration also allows to reduce the emission of greenhouse gases (particularly CO2 emission). The production of electricity being on-site, the burden on the utility network is reduced and the transmission line losses eliminated.

6 0

3 years ago

Read 2 more answers