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3 years ago

How much H2 is generated from the electrolysis of 150 grams of H2O

Chemistry

1 answer:

Romashka [77]3 years ago

5 0

Answer:

16.67 grams of H₂ is generated from the electrolysis of 150 grams of H₂O

Explanation:

Electrolysis is the decomposition of a chemical element under the effect of an electric current. So, electrolysis of water is the process of decomposing the H₂O molecule into separate oxygen and hydrogen gases due to an electric current passing through the water.

The balanced equation of electrolysis of water is:

2 H₂O → O₂ + 2H₂

Being:

H: 1 g/mole
O: 16 g/mole

then the molar mass of the compounds that participate in the reaction is:

H₂O: 2*1 g/mole + 16 g/mole= 18 g/mole
H₂: 2*1 g/mole= 2 g/mole
O₂: 2*16 g/mole= 32 g/mole

If the following amounts in moles are reacted by stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction):

H₂O: 2 moles
H₂: 2 moles
O₂: 1 mole

the amount of mass, by stoichiometry, that reacts and is produced is:

H₂O: 2 moles*18 g/mole=36 g
H₂: 2 moles* 2 g/mole= 4 g
O₂: 1 mole* 32 g/mole= 32 g

Then you can apply the following rule of three: if by stoichiometry 36 g of H₂O generate 4 g of H₂, 150 g of H₂O how much mass of H₂ will it generate?

$massofH_{2} =\frac{150 grams of H_{2}O*4 grams ofH_{2} }{36 grams of H_{2}O}$

mass of H₂= 16.67 grams

16.67 grams of H₂ is generated from the electrolysis of 150 grams of H₂O

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Lana71 [14]

Answer:

Oxide of M is $M_2O_3$ and sulfate of $M_2(SO_4)_3$

Explanation:

0.303 L of molecular hydrogen gas measured at 17°C and 741 mmHg.

Let moles of hydrogen gas be n.

Temperature of the gas ,T= 17°C =290 K

Pressure of the gas ,P= 741 mmHg= 0.9633 atm

Volume occupied by gas , V = 0.303 L

Using an ideal gas equation:

$PV=nRT$

$n=\frac{PV}{RT}=\frac{0.9633 atm\times 0.303 L}{0.0821 atm L/mol K\times 290 K}=0.01225 mol$

Moles of hydrogen gas produced = 0.01225 mol

$2M+2xHCl\rightarrow 2MCl_x+xH_2$

Moles of metal = $\frac{0.225 g}{27.0 g/mol}=8.3333 mol$

So, 8.3333 mol of metal M gives 0.01225 mol of hydrogen gas.

$\frac{8.3333}{0.01225 mol}=\frac{2}{x}$

x = 2.9 ≈ 3

$2M+6HCl\rightarrow 2MCl_3+3H_2$

$MCl_3\rightarrow M^{3+}+Cl^-$

Formulas for the oxide and sulfate of M will be:

Oxide of M is $M_2O_3$ and sulfate of $M_2(SO_4)_3$ .

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3 years ago

A bomb calorimeter has a heat capacity of 783 J/oC and contains 254 g of water whose specific heat capacity is 4.184 J/goC. How

IrinaK [193]

Answer : The amount of heat evolved by a reaction is, 4.81 kJ

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the water

$q=[q_1+q_2]$

$q=[c_1\times \Delta T+m_2\times c_2\times \Delta T]$

where,

q = heat released by the reaction

$q_1$ = heat absorbed by the calorimeter

$q_2$ = heat absorbed by the water

$c_1$ = specific heat of calorimeter = $783J/^oC$

$c_2$ = specific heat of water = $4.184J/g^oC$

$m_2$ = mass of water = 254 g

$\Delta T$ = change in temperature = $T_2-T_1=(23.73-26.01)=-2.28^oC$

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

$q=[(783J/^oC\times -2.28^oC)+(254g\times 4.184J/g^oC\times -2.28^oC)]$

$q=-4208.28J=-4.81kJ$

Therefore, the amount of heat evolved by a reaction is, 4.81 kJ

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Answer:

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