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3 years ago

How many moles of sodium are present in 17 g of Na?

Chemistry

1 answer:

11Alexandr11 [23.1K]3 years ago

8 0

Answer:

1. 0.74mol

2. 0.42mol

3. 2.125mol

4. 0.301mol

5. 4.52 × 10^23 particles

Explanation:

Number of moles (n) in a substance can be found using the formula:

mole (n) = mass/molar mass

Using this formula, the following moles are calculated:

1. Molar of Na = 23g/mol

mole = 17/23

mole = 0.74mol

2. Molar mass of Na2SO4 = 23(2) + 32 + 16(4)

= 46 + 32 + 64

= 142g/mol

Mole = 60/142

mole = 0.42mol

3. Molar mass of CO2 = 12 + 16(2)

= 12 + 32

= 44g/mol

mole = 93.5/44

mole = 2.125mol

4. Molar mass of sodium nitrate (NaNO3) = 23 + 14 + 16(3)

= 23 + 14 + 48

= 85g/mol

mole = 25.6/85

mole = 0.301mol

5. Number of particles in one mole of a substance is 6.022 × 10^23 particles. Hence, in 0.75mol of calcium hydroxide (Ca(OH)2, there will be;

0.75mol × 6.02 × 10^23

= 4.515 × 10^23

= 4.52 × 10^23 particles

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Write the balanced reaction using the fewest whole number coefficients to describe the reaction between gaseous hydrogen and gas

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Answer: The percentage yield of water is 9.5 %

Explanation:

We are given:

Moles of hydrogen = 14 moles

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By Stoichiometry of the reaction:

2 moles of hydrogen gas reacts with 1 mole of oxygen gas

So, 14 moles of hydrogen gas will react with = $\frac{1}{2}\times 14=7mol$ of oxygen gas

As, given amount of oxygen gas is more than the required amount. So, it is considered as an excess reagent

Thus, hydrogen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of hydrogen gas produces 2 moles of water

So, 14 moles of hydrogen gas will produce = $\frac{1}{2}\times 14=7mol$ of water

To calculate the percentage yield of water, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of water = 1.33 moles

Theoretical yield of water = 14 moles

Putting values in above equation, we get:

$\%\text{ yield of water}=\frac{1.33mol}{14mol}\times 100\\\\\% \text{yield of water}=9.5\%$

Hence, the percent yield of the water is 9.5 %.

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3 years ago

What is the difference between atomic mass, relative atomic mass and average atomic mass

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$\bold{\huge{\underline{\purple{ Answer }}}}$

Difference between Atomic mass, relative atomic mass and average atomic mass :-

<h3>Atomic Mass :-</h3>

Atomic mass is the mass of neutrons and protons present in the nucleus of an atom .
It is always calculated for a single element and having direct value
For isotopes also, the atomic mass is calculated separately . Example :- Carbon 12 , carbon 13 and carbon 14 have different atomic mass. 
The SI unit of Atomic mass is " u" and "amu"

<h3>Relative Atomic mass :-</h3>

Relative atomic mass is mean mass of the atoms of an element which is compared to the 1/12th mass of carbon - 12 .
Carbon - 12 is taken as a relative when we calculate the relative atomic mass of any element
For calculating relative atomic mass, we need to know the masses, percentage and abundance of all types of elements
Relative atomic mass is a dimension less quantity

<h3>Average Atomic Mass :-</h3>

Average atomic mass is the average mass of an atoms of a particular element by considering it's isotopes
While we calculate average atomic mass is a standardized number. Whereas, Average atomic mass sometimes varies geologically .
It also includes percentage, abundance and masses of given element .
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Ammonia is produced from the reaction of nitrogen and hydrogen according to the following balanced equation:

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Answer: 1) Maximum mass of ammonia 198.57g

2) The element that would be completely consumed is the N2

3) Mass that would keep unremained, is the one of the excess Reactant, that means the H2 with 3,44g

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In order to calculate the Mass of ammonia , we first check the Equation is actually Balance:

N2(g) + 3H2(g) ⟶2NH3(g)

Both equal amount of atoms side to side.

Now we verify which reagent is the limiting one by comparing the amount of product formed with each reactant, and the one with the lowest number is the limiting reactant. ( Keep in mind that we use the molecular weight of 28.01 g/mol N2; 2.02 g/mol H2; 17.03g/mol NH3)

Moles of ammonia produced with 163.3g N2(g) ⟶ 163.3g N2(g) x (1mol N2(g)/ 28.01 g N2(g) )x (2 mol NH3(g) /1 mol N2(g)) = 11.66 mol NH3

Moles of ammonia produced with 38.77 g H2⟶ 38.77 g H2 x ( 1mol H2/ 2.02 g H2 ) x (2 mol NH3 /3 mol H2 ) = 12.79 mol NH3

As we can see the amount of NH3 formed with the N2 is the lowest one , therefore the limiting reactant is the N2 that means, N2 is the element that would be completey consumed, and the maximum mass of ammonia will be produced from it.
We proceed calculating the maximum mass of NH3 from the 163.3g of N2.

11.66 mol NH3 x (17.03 g NH3 /1mol NH3) = 198.57 g NH3

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163.3g N2 x (1mol N2/28.01 g N2) x ( 3 mol H2 / 1 mol N2)x (2.02 g H2/ 1 mol H2) = 35.33 g H2

That means that only 35.33 g H2 will react with 163.3g N2 however we were giving 38.77g of H2, thus, 38.77g - 35.33 g = 3.44g H2 is left

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3 years ago