12.0 grams of sodium reacts with 5.00 grams of chlorine. What mass of sodium chloride could be produced? Na(s) + Cl2(g) à NaCl(s) (1) Identify the limiting reactant. (2) Determine the amount of sodium chloride produced.

Answers

Answer 1

Answer:

1. Cl₂ is the limiting

2. 8.24 g of NaCl.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

2Na + Cl₂ —> 2NaCl

Next, we shall determine the masses of Na and Cl₂ that reacted and the mass of NaCl produced from balanced equation. This is illustrated below:

Molar mass of Na = 23 g/mol

Mass of Na from the balanced equation = 2 × 23 = 46 g

Molar mass of Cl₂ = 2 × 35.5

= 71 g/mol

Mass of Cl₂ from the balanced equation = 1 × 71 = 71 g

Molar mass of NaCl = 23 + 35.5

= 58.5 g/mol

Mass of NaCl from the balanced equation = 2 × 58.5 = 117 g

SUMMARY:

From the balanced equation above,

46 g of Na reacted with 71 g of Cl₂ to produce 117 g of NaCl.

1. Determination of the limiting reactant.

From the balanced equation above,

46 g of Na reacted with 71 g of Cl₂.

Therefore, 12 g of Na will react with = (12 × 71) / 46 = 18.52 g of Cl₂.

From the calculation made above, we can see clearly that a higher amount (i.e 18.52 g) of Cl₂ than what was given (i.e 5 g) is needed to react completely with 12 g of Na.

Therefore, Cl₂ is the limiting reactant and Na is the excess reactant.

2. Determination of the amount of NaCl produced from the reaction.

In this case the limiting reactant will be used because it will give the maximum yield of NaCl as all of it consumed in the reaction.

The limiting reactant is Cl₂ and the mass of NaCl produced can be obtained as follow:

From the balanced equation above,

71 g of Cl₂ reacted to produce 117 g of NaCl.

Therefore, 5 g of Cl₂ will react to produce = (5 × 117)/71 = 8.24 g of NaCl.

Thus, 8.24 g of NaCl were obtained from the reaction.


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Answer:

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Answers

Convert all the gases from grams to moles using their molar masses first. Remember that nitrogen and oxygen exist as DIATOMIC gases.

O2 - 25 g / 32 g/mol = 0.78 mol O2
N2 - 15 g / 28.02 g/mol = 0.54 mol N2
He - 10 g / 4.0 g/mol = 2.5 mol He

Add all the moles of gases.

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Divide O2 moles by total moles.

0.78/3.82 = 0.20

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What is mole fraction?

The mole fraction, often known as the molar fraction, is defined in chemistry as the amount of a constituent, nᵢ, divided by the total amount of all constituents in a mixture, nₜₒₜ. It denotes the number of moles of a certain component in the solution divided by the total number of moles in the solution.

In order to calculate the mole fraction, initially the moles of the substance to be calculated from the given mass.

Gram to mole conversion,

[tex]=Grams\;*\;\frac{1\;mol}{grams} \\Moles of O2,25\;g\;*\;\frac{1\;mol}{32\;g}\;=\;0.78\;mol\;O2\\Moles of N215\;g\;*\;\frac{1\;mol}{28.02\;g}\;=\;0.54\;mol\;O2\\Moles of He10\;g\;*\;\frac{1\;mol}{4.0\;g}\;=\;2.5\;mol\;He[/tex]

Now, adding all the moles of gases0.78 + 0.54 + 2.5 = 3.82 moles.

Mole fraction of oxygen comes out to be,

[tex]Mole\; fraction\;=\frac{Amount \;of\;the\;substance}\\{Total\;moles} \\Mole\;fraction\;=\frac{0.78\;mol\;O2}{3.82 \;mol} =0.204.[/tex]

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Answers

Answer:

0.559 L

Explanation:

Step 1: Given data

Moles of sodium iodide (n): 0.405 molMolar concentration of sodium iodide (M): 0.724 M (0.724 mol/L)

Step 2: Calculate the volume of solution (V)

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The volume in Litres of a 0.724 M Nal solution that contains 0.405 mol of Nal is 0.56L.

HOW TO CALCULATE VOLUME:

The volume of a solution can be calculated by dividing the number of moles by the molarity as follows:

Volume = no. of moles ÷ molarity

According to this question, 0.724 M sodium iodide (NaI) solution contains 0.405 mol of Nal, the volume can be calculated as follows:

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Answers

Answer:

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Explanation:

NOTE: Since no hydrogen is attached to the compound given in question above, it means the compound is CCl₄.

The number of mole present in 78.2 g of CCl₄ can be obtained as follow:

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Answer:

hope it helps you

mark me as brainliest!!

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24.50 g of Na react with H2O to make sodium hydroxide and H, gas. How much sodium
hydroxide is made if there is an excess of water?

Answers

Answer:

42.61 g of NaOH.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

2Na + 2H₂O —> 2NaOH + H₂

Next, we shall determine the mass of Na that reacted and the mass of NaOH produced from the balanced equation. This is illustrated below:

Molar mass of Na = 23 g/mol

Mass of Na from the balanced equation = 2 × 23 = 46 g

Molar mass of NaOH = 23 + 16 + 1

= 40 g/mol

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SUMMARY:

From the balanced equation above,

46 g of Na reacted to produce 80 g of NaOH.

Finally, we shall determine the mass of NaOH produced by the reaction of 24.50 g of Na. This can be obtained as follow:

From the balanced equation above,

46 g of Na reacted to produce 80 g of NaOH.

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