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3 years ago

How many moles of hydrogen gas would be needed to react with excess carbon dioxide to produce 65.1 moles of water vapor?

Chemistry

2 answers:

ivolga24 [154]3 years ago

5 0

CO2 + 4 H2 -----> CH4 + 2H2O ......

Likurg_2 [28]3 years ago

4 0

The relevant chemical reaction can be written as:

In this reaction, carbon dioxide reacts with hydrogen to form methane and water. This reaction is known as the Sabatier reaction and is very commonly used to generate methane for fuel.

Using stoichiometry, along with the balanced chemical reaction (written above), 4 moles of hydrogen gas are used to produce 2 moles of water.

or, 2 moles of hydrogen will produce 1 mole of water.

Thus, to produce 88.1 moles of water, 2 x 88.1 = 176.2 moles of hydrogen gas would be needed in the presence of excess carbon dioxide. Knowing this data, we can also calculate the mass of hydrogen gas needed. The molar mass of hydrogen gas (H_2) is 2 g/mole and hence a total of 352.4 grams of hydrogen gas are needed.

Hope this helps! :3

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The difference in the heights of the Hg columns in Diagram B is17 mm-Hg. If the atmospheric pressure is 769 mm-Hg, what is the p

Korvikt [17]

The pressure of the confined gas will be 769-17 = 752 mmHg

<h3>What is Manometer ?</h3>

Manometer is a U shaped instrument which is used to measure the pressure acting on a column of fluid , a difference in the pressure in the two arms of the manometer causes the liquid to reach different heights.

The difference in the heights of the Hg columns in Diagram B is17 mm-Hg.

the atmospheric pressure is 769 mm-Hg,

the pressure of the confined gas in mm-Hg = ?

The pressure of the confined gas will be 769-17 = 752 mmHg

To know more about Manometer

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3 0

1 year ago

How many chlorine atoms are there in 12.2 g of ccl4? How many chlorine atoms are there in 12.2 of ? 4.77×1022 atoms?

mash [69]

Answer:- $1.91*10^2^3Cl$ atoms.

Solution:- We have been given the grams of carbon tetrachloride and asked to calculate the number of atoms of chlorine. It is a three step conversion problem. In the first we convert the grams of carbon tetrachloride to moles of it. In second step we convert moles of carbon tetrachloride to moles of chlorine and in the third step we convert the moles of chlorine to atoms of chlorine.

For grams to mole conversion we need the molar mass of the compound. Molar mass of carbon tetrachloride is 153.82 grams per mol. If we look at the formula of carbon tetrachloride then four chlorine are present in it. It means 1 mol of carbon tetrachloride has four moles of chlorine. The calculations are as follows:

$12.2gCCl_4(\frac{1molCCl_4}{153.82gCCl_4})(\frac{4molCl}{1molCCl_4})(\frac{6.02*10^2^3Clatoms}{1molCl})$

= $1.91*10^2^3Cl$ atoms

So, there are $1.91*10^2^3Cl$ atoms in 12.2 grams of $CCl_4$ .

6 0

3 years ago

Read 2 more answers

"the charcoal from a tree killed in the volcanic eruption that formed cedar lake in oregon contained 44.5% of the carbon-14 foun

nalin [4]

The half life of carbon-14 is 5700 years. (Half life is the time taken by a radioactive isotope to decay by half of its original mass).
Let A₀ be the initial amount of carbon-14 that is found in living matter (t=0 years), to determine when there was 44.5% of A₀ left.
44.5 = 100 × (1/2)^n, where n is the number of half lives
0.5^n = 0.445
n = log 0.445/log 0.5
n = 1.168
But 1 half life is 5700 years
Therefore, the number of years will be 5700 × 1.168 = 6658.299725 years
≈ 6658.30 years

3 0

2 years ago

Using the balanced equation N2+O2=2NO, how many grams of NO can be produced when 25.0 grams of N react?

Oxana [17]

Answer:

$53.55gNO$

Explanation:

Hello there!

In this case, according to the given chemical reaction, it is possible for us to calculate the produced grams of nitrogen monoxide by starting with 25.0 g of nitrogen via their 1:2 mole ratio and the molar masses of 30.1 g/mol and 28.02 g/mol, respectively and by some stoichiometry:

$=25.0gN_2*\frac{1molN_2}{28.02gN_2}*\frac{2molNO}{1molN_2}*\frac{30.01 gNO}{1molNO}\\\\=53.55gNO$

Best regards!

8 0

3 years ago

A power plant is driven by the combustion of a complex fossil fuel having the formula C11H7S. Assume the air supply is composed

AlekseyPX

(a) 4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 20.68N_2;

(b) 4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2;

(a) Balanced equation including N_2 from air

The balanced equation ignoring N_2 from air is

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2

Moles of N_2 =55 mol O_2 × (3.76 mol N_2/1 mol O_2) = 206.8 mol N_2

Including N_2 from air, the balanced equation is

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 206.8N_2

(b) Balanced equation for 120 % stoichiometric combustion

Moles of O_2 = 55 mol O_2 × 1.20 = 66.00 mol O_2

Excess moles O_2 = (66.00 – 55) mol O_2 = 11.00 mol O_2

Moles of N_2 = 66.00 mol O_2 × (3.76 mol N_2/1 mol O_2) = 248.2 mol N_2

The balanced equation is

4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2

Moles of O_2 required = 1700 kg C_11 H_7S

× (1 kmol C_11 H_7S/185.24 kg C_11 H_7S) × (55 kmol O_2/4 kmol C_11 H_7S)

= 126.2 kmol O_2

Mass of O_2 = 126.2 kmol O_2 × (32.00 kg O_2/1 kmol O_2) = 4038 kg O_2

Mass of N_2 required = 126.2 kmol O_2 × (3.76 kmol N_2/1 kmol O_2)

× (28.01 kg N_2/1 kmol N_2) = 13 285 kg N_2

Mass of air = Mass of N_2 + mass of O_2 = (4038 + 13 285) kg = 17 300 kg air

(d) Air:fuel mass ratio for 100 % combustion

Air:fuel = 17 300 kg/1700 kg = 10.2 :1

(e) Air:fuel mass ratio for 120 % combustion

Mass of air = 17 300 kg × 1.20 = 20 760 kg air

Air:fuel = 20 760 kg/1700 kg = 12.2 :1

6 0

2 years ago