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2 years ago

How many grams of Al would be needed to displace all of the copper ions in 71.8 mL of 0.610 M CuSO4?

Chemistry

1 answer:

Galina-37 [17]2 years ago

4 0

The amount of Al that would be needed will be 0.79 grams

<h3>Stoichiometric calculations</h3>

From the equation of the reaction below:

$2 Al + 3 CuSO_4 --- > Al_2(SO_4)_3 + 3 Cu$

The mole ratio of Al to $CuSO_4$ is 2:3.

Mole of 71.8 mL. 0.610 M $CuSO_4$ = 0.610 x 71.8/1000 = 0.0438 moles

Equivalent mole of Al = 2/3 x 0.0438 = 0.029 moles

Mass of o.o29 moles Al = 0.029 x 26.98 = 0.79 grams

More on stoichiometric calculations can be found here: brainly.com/question/27287858

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How many grams are in 2.30 x 10^23 atoms in Na?

aksik [14]

There are 8.786 g

<h3>Further explanation </h3>

The mole is the number of particles contained in a substance

1 mol = 6.02.10²³

2.30 x 10²³ atoms in Na :

$\tt mol=\dfrac{2.3\times 10^{23}}{6.02\times 10^{23}}=0.382$

mass :

$\tt mass=mol\times Ar~Na\\\\mass=0.382\times 23=8.786~g$

4 0

3 years ago

How many grams of oxygen will be required for sulfur to be burned to give 100.0 g of so2

klemol [59]

Answer:

50.05 g.

Explanation

you welcome ;)

4 0

3 years ago

How many grams of carbon are in 36.6 grams in glucose

Lana71 [14]

Answer:

14.64 grams of carbon

Explanation:

The Molecular formula of Glucose is C₆H₁₂O₆.

From the molecular formula, the Molar mass of Glucose is,

Mass of Carbon in Glucose = 6*12 = 72.

Mass of Hydrogen in Glucose = 12*1 = 12.

Mass of Oxygen in Glucose = 6*16 = 96.

Molar mass of Glucose = 72+12+96 = 180 grams.

180 grams of Glucose contains 72 grams of Carbon.

How many grams of Carbon are there in 36.6 grams of Glucose.

180 --> 72

36.6 --> ?

Let it be 'x' grams

Then, by Criss Cross, (x)(180) = (36.6)(72) => x = 14.64 grams.

Therefore 14.64 grams of carbon are there in 36.6 grams in glucose.

8 0

3 years ago

For the following reaction, 5.61 grams of carbon monoxide are mixed with excess water . Assume that the percent yield of carbon

stealth61 [152]

<u>Answer:</u> The ideal yield of carbon dioxide is 7.506 grams

<u>Explanation:</u>

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 carbon monoxide = 5.61 g

Molar mass of carbon monoxide = 28 g/mol

Putting values in equation 1, we get:

$\text{Moles of carbon monoxide}=\frac{5.61g}{28g/mol}=0.200mol$

The chemical equation for the reaction of carbon monoxide and water follows:

$CO(g)+H_2O(l)\rightarrow CO_2(g)+H_2(g)$

By Stoichiometry of the reaction:

1 mole of carbon monoxide produces 1 mole of carbon dioxide

So, 0.200 moles of carbon monoxide will produce = $\frac{1}{1}\times 0.200=0.200mol$ of carbon dioxide

Now, calculating the mass of carbon dioxide from equation 1, we get:

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = 0.200 moles

Putting values in equation 1, we get:

$0.200mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=(0.200mol\times 44g/mol)=8.8g$

To calculate the experimental yield of carbon dioxide, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Percentage yield of carbon dioxide = 85.3 %

Theoretical yield of carbon dioxide = 8.8 g

Putting values in above equation, we get:

$85.3=\frac{\text{Experimental yield of carbon dioxide}}{8.8g}\times 100\\\\\text{Experimental yield of carbon dioxide}=\frac{85.3\times 8.8}{100}=7.506g$

Hence, the ideal yield of carbon dioxide is 7.506 grams

3 0

3 years ago

In an automobile's catalytic converter,

tia_tia [17]

The most of the pollutants (carbon monoxide, nitrogen monoxide and unburned hydrocarbons) are released from the car during the first one to two minutes after the car is started, and then decrease significantly after that, due to increasing the temperature which supply the system by heat that required to oxidize the released gases by increasing the rate of the chemical reaction.

Then, the correct answer is the temperature (B).

5 0

3 years ago