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3 years ago

What mass of Cu(s) would be produced if 0.40 mol of Cu2O(s) was reduced completely with excess H2(g) ?

Chemistry

1 answer:

yan [13]3 years ago

4 0

Answer:

50.8g

Explanation:

Given parameters:

Nubmer of mole of Cu₂O = 0.4mol

Unknown:

Mass of Cu produced = ?

Solution:

The balanced reaction equation:

Cu₂O + H₂ ⇄ 2Cu + H₂O

From the reaction, we know that:

1 mole of Cu₂O produced 2 moles of Cu

0.4 mol of Cu₂O would give 0.8mol of Cu

Now we know the number of moles of Cu produced because the limiting reactant here is Cu₂O. It was used up in the reaction in which the hydrogen gas was in excess.

To find the mass of Cu produced, we use the equation below :

Mass of Cu = Number of moles of Cu x molar mass of Cu

Molar mass of Cu = 63.5gmol⁻¹

Mass of Cu = 0.8 x 63.5 = 50.8g

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The frequency factors for these two reactions are very close to each other in value. Assuming that they are the same, compute th

MrRissso [65]

The question is incomplete, complete question is :

The frequency factors for these two reactions are very close to each other in value. Assuming that they are the same, compute the ratio of the reaction rate constants for these two reactions at 25°C.

$\frac{K_1}{K_2}=?$

Activation energy of the reaction 1 , $Ea_1$ = 14.0 kJ/mol

Activation energy of the reaction 2, $Ea_1$ = 11.9 kJ/mol

Answer:

0.4284 is the ratio of the rate constants.

Explanation:

According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

The expression used with catalyst and without catalyst is,

$\frac{K_2}{K_1}=\frac{A\times e^{\frac{-Ea_2}{RT}}}{A\times e^{\frac{-Ea_1}{RT}}}$

$\frac{K_2}{K_1}=e^{\frac{Ea_1-Ea_2}{RT}}$

where,

$K_2$ = rate constant reaction -1

$K_1$ = rate constant reaction -2

Activation energy of the reaction 1 , $Ea_1$ = 14.0 kJ/mol = 14,000 J

Activation energy of the reaction 2, $Ea_1$ = 11.9 kJ/mol = 11,900 J

R = gas constant = 8.314 J/ mol K

T = temperature = $25^oC=273+25=298 K$

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

$\frac{K_1}{K_2}=e^{\frac{11,900- 14,000Jl}{8.314 J/mol K\times 298 K}}=2.3340$

0.4284 is the ratio of the rate constants.

7 0

3 years ago

How many grams of hydrochloric acid are produced when 15.0 grams NaCl react with excess H2SO4 in the reaction

e-lub [12.9K]

Answer:

11.65 g

Explanation:

Amount of NaCl = 15.0 grams

amount of H₂SO₄ = Excess

mass of hydrochloric acid (HCl) = ?

Solution:

To solve this problem first we will look for the reaction that NaCl react with H₂SO₄

Reaction:

2NaCl + H₂SO₄ --------> 2 HCl + Na₂SO₄

As the H₂SO₄ is in excess so the amount of hydrochloric acid (HCl) depends on the amount of NaCl as it its act as limiting reactant.

Now if we look at the reaction

2NaCl + H₂SO₄ --------> 2HCl + Na₂SO₄

1 mol 2 mol

Now convert moles to mass

Then

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

Molar mass of HCl = 1 + 35.5 = 36.5 g/mol

So,

2NaCl + H₂SO₄ --------> 2HCl + Na₂SO₄

2 mol (58.5 g/mol) 2 mol (36.5 g/mol)

94 g 73 g

So if we look at the reaction; 94 g of NaCl gives with 73 g of hydrochloric acid (HCl) in a this reaction, then how many grams of hydrochloric acid (HCl) will produce from 15.0 g of NaCl

For this apply unity formula

94 g of NaCl ≅ 73 g of HCl

15.0 g of NaCl ≅ X g of HCl

By Doing cross multiplication

X g of HCl = 73 g x 15.0 g / 94 g

X g of HCl = 11.65 g

11.65 g of hydrochloric acid (HCl) will produce by 15.0 g of NaCl

7 0

3 years ago

He chemical formula for lithium hydride is LiH A chemist determined by measurements that 0.0800 moles of lithium hydride partici

sweet-ann [11.9K]

Answer:

Mark me brainlist I will then answer in comment

3 0

3 years ago

I really need help on this question please help

satela [25.4K]

Answer:

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Hope this helped ! :)

6 0

3 years ago

two sides of a triangle have the following measures . Find The range of possible measures for the tird side

crimeas [40]

To find the third side you would use Pythagorean theorem which is a²+b²=c².

A and B being the 2 legs and C being the hypotenuse.

Example: ( sorry for the really odd not really a triangle, triangle...)
| \
6| \ ? 6²+9²=C²
|_______ 36+81=C²
9 117=C²
√117=10.8 (rounded)

if you need to get the leg you just fill in the numbers.

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3 years ago