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3 years ago

How many grams of water are made from the reaction of 4.0 grams of hydrogen gas with the given reaction 2H2+O2--->2H2O

Chemistry

1 answer:

Sergeeva-Olga [200]3 years ago

4 0

Answer:

The mass of water is 36 g.

Explanation:

Mass of hydrogen = 4 g

Mass of water = ?

Solution:

First of all we will write the balance chemical equation:

2H₂ + O₂ → 2H₂O

Number of moles of hydrogen = mass / molar mass

Number of moles of hydrogen = 4 g/ 2 g/mol

Number of moles of hydrogen = 2 mol

Now we compare the moles of water with hydrogen from balance chemical equation.

H₂ : H₂O

2 : 2

Mass of water = moles × molar mass

Mass of water = 2 mol × 18 g/mol

Mass of water = 36 g

If the water oxygen is in excess than mass of water would be 36 g.

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Five kg of carbon dioxide (CO2) gas undergoes a process in a well-insulated piston-cylinder assembly from 2 bar, 280 K to 20 bar

kaheart [24]

This question is incomplete, the complete question is;

Five kg of carbon dioxide (CO2) gas undergoes a process in a well-insulated piston-cylinder assembly from 2 bar, 280 K to 20 bar, 520 K. If the carbon dioxide behaves as an ideal gas, determine the amount of entropy produced, in kJ/K. Assuming;

a) constant specific heats Cp = 0.939 kJ/Kg K

b) variable specific heats

Answer:

a) the amount of entropy produced is 0.731599 kJ/K

b) the amount of entropy produced is 0.69845 kJ/K

Explanation:

Given the data in the question;

5 kg of carbon dioxide (CO₂) gas undergoes a process in a well-insulated piston-cylinder assembly.

m = 5 kg

Molar mass M = 44.01 g/mol

P₁ = 2 bar, P₂ = 20

T₁ = 280 K, P₂ = 520 K

Since its insulated { q = 0 } ( kinetic and potential energy effects = 0 )

Now,

a) the amount of entropy produced, in kJ/K, Assuming constant specific heats with Cp = 0.939 kJ/Kg K

S $_{Generation$ = m × ((Cp × In( T₂/T₁) - R × In( P₂/p₁ ))

we substitute

S $_{Generation$ = 5 × (( 0.939 × In( 520/280) - 0.1889 × In( 20/2 ))

= 5 × ( 0.5812778 - 0.434958 )

= 5 × 0.1463198

= 0.731599 kJ/K

Therefore, the amount of entropy produced is 0.731599 kJ/K

b) the amount of entropy produced, in kJ/K, Assuming variable specific heats.

Now, from Table A-23: Ideal Gas Properties of Selected Gases;

T₁,T₂ : s₁⁰ = 211.376 kJ/kmol-K, s₂⁰ = 236.575 kJ/kmol-K

now, s₁ = s₁⁰ / M and s₂ = s₂⁰ / M

we substitute

s₁ = s₁⁰ / M = 211.376 / 44.01 = 4.8029 kJ/kg

s₂ = s₂⁰ / M = 236.575 / 44.01 = 5.37548 kJ/kg

S $_{Generation$ = m × (( s₂ - s₁ ) - R × In( p₂ / p₁ ))

we substitute

S $_{Generation$ = 5 × (( 5.37548 - 4.8029 ) - 0.1880 × In( 20 / 2 ))

= 5 × ( 0.57258 - 0.432885997 )

= 5 × 0.13969

= 0.69845 kJ/K

Therefore, the amount of entropy produced is 0.69845 kJ/K

5 0

3 years ago

How many moles of mgs2o3 are in 193 g of the compound?

Mashutka [201]

Answer:

1.414 Moles

Solution:

Data Given:

Mass of MgS₂O₃ = 193 g

M.Mass of MgS₂O₃ = 136.43 g.mol⁻¹

Moles = ?

Formula Used:

Moles = Mass ÷ M.Mass

Putting values,

Moles = 193 g ÷ 136.43 g.mol⁻¹

Moles = 1.414 mol

8 0

3 years ago

If a pork roast must absorb kJ to fully cook, and if only 10% of the heat produced by the barbecue is actually absorbed by the r

kolezko [41]

Answer:

1,033.56 grams of carbon dioxide was emitted into the atmosphere.

Explanation:

$C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g),\Delta H_{rxn}=-2044 kJ/mol (assuming)$

Energy absorbed by pork,E = $1.6\times 10^3 kJ$ (assuming)

Total energy produced by barbecue = Q

Percentage of energy absorbed by pork = 10%

$10\%=\frac{E}{Q}\times 100$

$Q=\frac{E}{10}\times 100=\frac{1.6\times 10^3 kJ}{10}\times 100=1.6\times 10^4 kJ$

Since, it is a energy produced in order to indicate the direction of heat produced we will use negative sign.

Q = $-1.6\times 10^4 kJ$

Moles of propane burnt to produce Q energy =n

$n\times \Delta H_{rxn}=Q$

$n=\frac{Q}{\Delta H_{rxn}}=\frac{-1.6\times 10^4 kJ}{-2044 kJ/mol}=7.83 mol$

According to reaction , 1 mol of propane gives 3 moles of carbon dioxide. then 7.83 moles of will give:

$\frac{1}{3}\times 7.83 mol=23.49 mol$ carbon dioxide gas.

Mass of 23.49 moles of carbon dioxide gas:

23.49 mol × 44 g/mol =1,033.56 g

1,033.56 grams of carbon dioxide was emitted into the atmosphere.

8 0

3 years ago

An element X combines with oxygen to form a compound of formula XO2. If 24.0 grams of element X combines with exactly 16.0 grams

tensa zangetsu [6.8K]

Answer:

atomic mass of X is 48.0 amu

Explanation:

Let y be the atomic mass of X

Molar mass of O_2 is = 2×16 = 32 g / mol

X + O2 -----> XO_2

According to the equation ,

y g of X reacts with 32 g of O_2

24 g of X reacts with Z g of O_2

Z = ( 32×24) / y

But given that 24.0 g of X exactly reacts with 16.0 g of O_2

So Z = 16.0

⇒ (32×24) / y = 16.0

⇒ y = (32×24) / 16

y= 48.0

So atomic mass of X is 48.0 amu

4 0

3 years ago

An atom that has 13 protons and 15 neutrons is isotope of the element A.) nickel B.) silicon C.) aluminum D.) phosphorus. Why?

joja [24]

An atom that has 13 protons and 15 neutrons is isotope of Aluminium (answer C)

Explanation

Isotope is a form of the same element with the equal number of protons but difference number of neutrons in their nuclei.

In other words isotope has the same atomic number but different mass number.

Atomic number of a element is determined by number of protons of an element.

from the periodic table Aluminum in atomic number 13 therefore it has 13 protons therefore an atom that has 13 protons and 15 neutrons is a isotope of Aluminium.

6 0

3 years ago

Read 2 more answers