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LuckyWell [14K]
2 years ago
6

2.3g of sodium metal is burned completely in oxygen of air.

Chemistry
1 answer:
Nina [5.8K]2 years ago
4 0

Answer:

mass of Na2O= 3.1 grams

volume of O2= 0.56 L

Explanation:

First what you need to do is establish the balanced equation for the reaction that they're giving you. In this case would be:

4Na + O2 > 2Na2O

And then, you can read the stequiometry of the reaction:

4 moles of Na react with one mole of oxygen to produce 2 moles of sodium oxide.

So now you know that you have 2.3 grams of sodium so you have to obtain the quantity of moles that are present in 2.3 grams of sodium. The molar mass the (the grams by which you will have exactly one mole) so all you have to do is divide 2.3g ÷ 23g/mol= 0.1 mol

So here you have 0.1 mol of sodium, and you know for a fact that for each 4 moles of sodium, one mole of oxygen reacts, so you simply divide the quantity of moles of sodium you have by the 4 moles of oxygen present in the balanced equation.

0.1 moles ÷ 4 moles = 0.025.

So for each 0.1 moles of sodium that react, 0.025 moles of Oxygen react.

Now what you need to do is to find the limitant reactor (the reactor that's according to the comparison to the balanced equation is in a smaller ratio to the other element, that would be in excess.

In this case, both elements are in a ratio that respects the balanced equation so neither of them is in excess or is limitant.

Now, to determine the amount of Na2O, you simply know that one mole of oxygen react to produce to moles of sodium oxide, So all you have to do is establish the factor of conversion:

0.025 moles O2 x 2 moles Na2O/ 1 mol O2= 0.05 moles of Na2O

Then you convert that to grams by multiplying 0.05 by the molar mass of Sodium oxide and it will be:

0.05 mol Na2O x 62 g/mol= 3.1 grams of Na2O

Now what you need to do is find the volume by which O2 is reacting in the reaction so you can use the equation of the ideal gas law.

PV=nRT at standard temperature and pressure (273 K and 1 Atm)

P=Pressure

V=Volume

n=Number of atoms

R=constant of gases (0.082 Atm L/ mol K)

T=Temperature

V=nRT/P (because you need to find the volume)

V= (0.025 mol)(0.082)(273 K)/1 atm

V=0.56 L

And there you go :)

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3 years ago
A 2.684-g sample of zinc oxide was reduced by hydrogen gas, resulting in 2. 156 g of pure zinc metal. Determine the empirical fo
Afina-wow [57]

The empirical formula of the initial zinc oxide is ZnO.

<h3>What is Empirical Formula?</h3>

The empirical formula of a compound represents the ratios of elements in a compound but not the actual numbers or arrangement of the atoms.

It is the lowest whole number ratio of the element in the compound.

<h3>How to find out the empirical formula?</h3>
  • Find out the given masses and molar masses of the elements

The molar mass of Zn = 65 gmol⁻¹

Given the mass of Zn = 2.156 g

The molar mass of Oxygen = 16 gmol⁻¹

The mass of Oxygen = Mass of a sample of zinc oxide - the mass of zinc metal

                                   = (2.684 - 2.156) g

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  • Find the number of moles of the elements in the compound

The number of moles is given by

n = \frac{m}{M}

where m = given mass and

M = Molar mass

Number of moles of Zinc = \frac{2.156}{65} = 0.033 moles

Number of moles of Oxygen =\frac{0.528}{16} = 0.033 moles

  • Find the simplest ratios of the elements in the compound. To find the ratios simply divide the number of moles by the lowest number of moles obtained.

Here, the number of moles is the same for both elements. Hence, the simplest ratio for Zn:O is 1:1.

Therefore, the empirical formula of zinc oxide is ZnO.

Learn more about the empirical formula:

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5 0
1 year ago
Write the appropriate symbol for each of the following isotopes: (a) Z 11, A 23; (b) Z= 28, A= 64; (c) Z= 50, A =115; (d) Z= 20,
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Explanation:  

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a) Z = 11, A =  23

Element = Sodium

  symbol: ²³₁₁Na  .

b) Z = 28, A =  64

Element = Nickel

  symbol: ⁶⁴₂₈Ni  .

c) Z = 50, A = 115

Element = tin

  symbol: ¹¹⁵₅₀Sn  .

d) Z = 20, A = 42

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  symbol: ⁴²₂₀Ca .

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The replacing of sodium hydroxide with potassium hydroxide (KOH) to the reaction will least affect the organic product that forms.

 

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The final temperature is 407.57 K

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