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3 years ago

C. If 150 grams of oxygen are produced, how many moles of sodium chlorate decomposed

Chemistry

1 answer:

Black_prince [1.1K]3 years ago

7 0

Answer:

3.13 moles of sodium chlorate are decomposed.

Explanation:

Given data:

Mass of oxygen produced = 150 g

Number of moles of sodium chlorate decomposed = ?

Solution:

Chemical equation:

2NaClO₃ → 2NaCl + 3O₂

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 150 g/ 32 g/mol

Number of moles = 4.7 mol

Now we will compare the moles of oxygen with sodium chlorate.

O₂ : NaClO₃

3 : 2

4.7 : 2/3×4.7 = 3.13 mol

3.13 moles of sodium chlorate are decomposed.

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Nitrogen dioxide is produced by combustion in an automobile engine. For the following reaction, 0.377 moles of nitrogen monoxide

Contact [7]

Answer:

The amount of NO₂ that can be produced 8.533 g

Explanation:

According to question

2 NO(g) + O₂(g) → 2 NO₂(g)

Given

Moles of nitrogen monoxide = 0.377

Moles of oxygen = 0.278

$'For NO'=\frac{Mole}{Stoichiometry}=\frac{0.377}{2} =0.1855\\'For O_{2} '=\frac{0.278}{1}= 0.278\\$

Since 'NO' is the limiting reagent according to this ratio.

According to equation

2 moles NO reacts to form 2 moles NO₂

So, 0.1855 moles NO give = 0.1855 moles of NO₂

Mass of 1 mole NO₂ = 46 g/mole

Mass of 0.1855 moles = 46 x 0.1855 = 8.533 g

5 0

3 years ago

What have six valence electrons

irina1246 [14]

Any element in group 18 has eight valence electrons (except for helium, which has a total of just two electrons). Examples include neon (Ne), argon (Ar), and krypton (Kr). Oxygen, like all the other elements in group 16, has six valence electrons.

6 0

3 years ago

Read 2 more answers

Carbon tetrachloride reacts at high temperatures with oxygen to produce two toxic gases, phosgene and chlorine. CCl4(g) + (1/2)O

madam [21]

The question is incomplete, here is the complete question:

Carbon tetrachloride reacts at high temperatures with oxygen to produce two toxic gases, phosgene and chlorine.

$CCl_4(g)+\frac{1}{2}O_2(g)\rightleftharpoons COCl_2(g)+Cl_2(g);K_c=4.4\times 10^9$ at 1,000 K

Calculate Kc for the reaction $2CCl_4(g)+O_2(g)\rightleftharpoons 2COCl_2(g)+2Cl_2(g)$

<u>Answer:</u> The value of $K_c'$ for the final reaction is $1.936\times 10^{19}$

<u>Explanation:</u>

The given chemical equations follows:

$CCl_4(g)+\frac{1}{2}O_2(g)\rightleftharpoons COCl_2(g)+Cl_2(g);K_c$

We need to calculate the equilibrium constant for the equation, which is:

$2CCl_4(g)+O_2(g)\rightleftharpoons 2COCl_2(g)+2Cl_2(g)$

As, the final reaction is the twice of the initial equation. So, the equilibrium constant for the final reaction will be the square of the initial equilibrium constant.

The value of equilibrium constant for net reaction is:

$K_c'=(K_c)^2$

We are given:

$K_c=4.4\times 10^9$

Putting values in above equation, we get:

$K_c'=(4.4\times 10^9)^2=1.936\times 10^{19}$

Hence, the value of $K_c'$ for the final reaction is $1.936\times 10^{19}$

3 0

3 years ago

A 10-gram aluminum cube absorbs 677 joules when its temperature is increased from 50°C to 125°C. What is the specific heat of al

ra1l [238]

<h3>Answer:</h3>

0.90J/g°C

<h3>Explanation:</h3>

We are given:

Mass of Aluminium = 10 g

Quantity of heat = 677 Joules

Change in temperature = 125°C - 50°C

= 75°C

We are required to calculate the specific heat capacity of Aluminium

But, Quantity of heat = Mass × specific heat × Change in temperature

Q = mcΔt

Rearranging the formula;

c = Q ÷ mΔt

= 677 J ÷ (10 g × 75°C)

= 677 J ÷ 750g°C

= 0.903 J/g°C

= 0.90J/g°C

Thus, the specific heat capacity of Aluminium is 0.90J/g°C

8 0

3 years ago

Zn +

saul85 [17]

Answer:

1. Theoretical yield = 2.03g

2. Actual yield 1.89g

Explanation:

Let us write a balanced equation. This is illustrated below:

Zn + 2HCI —> ZnCl2 + H2

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

Mass of HCl from the balanced equation = 2 x 36.5 = 73g

Molar Mass of H2 = 2x1 = 2g/mol

1. From the equation,

73g of HCl produced 2g of H2.

Therefore, 74g of HCl will produce = (74 x 2)/73 = 2.03g

Therefore, theoretical yield = 2.03g

2. %yield = 93%

Theoretical yield = 2.03g

Actual yield =?

%yield = Actual yield /Theoretical yield x100

Actual yield = %yield x theoretical yield

Actual yield = 93% x 2.03 = (93/100)x2.03 = 1.89g

Actual yield =1.89g

4 0

3 years ago