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3 years ago

A sample of 23.2 g of nitrogen gas is reacted with

Chemistry

1 answer:

slavikrds [6]3 years ago

5 0

Answer:

1.66 moles.

Explanation:

We'll begin by calculating the number of mole in 23.2 g of nitrogen gas, N2.

This is illustrated below:

Molar mass of N2 = 2x14 = 28 g/mol

Mass of N2 = 23.2 g

Mole of N2 =.?

Mole = mass /Molar mass

Mole of N2 = 23.2/28

Mole of N2 = 0.83 mole

Next, we shall determine the number of mole in 23.2 g of Hydrogen gas, H2.

This is illustrated below:

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

Mass of H2 = 23.2 g

Mole of H2 =?

Mole = mass /Molar mass

Mole of H2 = 23.2/2

Mole of H2 = 11.6 moles

Next, the balanced equation for the reaction. This is given below:

N2 + 3H2 —> 2NH3

From the balanced equation above,

1 mole of N2 reacted with 3 moles of H2 to produce 2 moles of NH3.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

1 mole of N2 reacted with 3 moles of H2.

Therefore, 0.83 moles will react with = (0.83 x 3) = 2.49 moles of H2.

From the calculations made above, we can see that only 2.49 moles out of 11.6 moles of H2 is required to react completely with 0.83 mole of N2.

Therefore, N2 is the limiting reactant.

Finally, we shall determine the maximum amount of NH3 produced from the reaction.

In this case, we shall use the limiting reactant because it will give the maximum yield of NH3 since all of it is consumed in the reaction.

The limiting reactant is N2 and the maximum amount of NH3 produced can be obtained as follow:

From the balanced equation above,

1 mole of N2 reacted to produce 2 moles of NH3.

Therefore, 0.83 mole of N2 will react to produce = (0.83 x 2) = 1.66 moles of NH3.

Therefore, the maximum amount of NH3 produced from the reaction is 1.66 moles.

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Describe a polyhydroxyl alcohol.

Mademuasel [1]

With a name like poly hydroxyl alcohol, it suggests that the alcohol has more than one alcohol group. Thus any alcohol with more than one hydroxyl is a polyhydroxyl alcohol. An example that pops into mind is ethanediol (CH2OHCH2OH) this has 2 hydroxyl groups and is an alcohol, thus a polyhydroxyl alcohol

8 0

3 years ago

A 100.0 mL solution containing 0.864 g of maleic acid (MW=116.072 g/mol) is titrated with 0.276 M KOH. Calculate the pH of the s

Lilit [14]

Answer:

pH = 1.32

Explanation:

H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺

This problem involves a weak diprotic acid which we can solve by realizing they amount to buffer solutions. In the first deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:

So first calculate the moles reacted and produced:

n H₂M = 0.864 g/mol x 1 mol/ 116.072 g = 0.074 mol H₂M

54 mL x 1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH

it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.

moles H₂M left = 0.074 - 0.015 = 0.059

moles HM⁻ produced = 0.015

Using the Henderson - Hasselbach equation to solve for pH:

ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325

Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.

For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.

3 0

3 years ago

You weigh out 0.1183 g of a complex salt to analyze for the percentage of cyanide ion in your complex salt. After dissolving the

Stels [109]

Answer:

25.35%

Explanation:

Again let me restate the the equation of the reaction;

H2O (ℓ) + 2 MnO4 - (aq) + 3 CN- (aq) → 2 MnO2 (s) + 3 CNO- (aq) + 2 OH- (aq)

Amount of potassium permanganate reacted = 10.2/1000 * 0.08035 = 8.1957 * 10^-4 moles

If 2 moles of MnO4 - reacts with 3 moles of CN-

8.1957 * 10^-4 moles of MnO4 - reacts with 8.1957 * 10^-4 * 3/2

= 1.229 * 10^-3 moles of CN-

Mass of CN- reacted = 1.229 * 10^-3 moles of CN- * 26.02 g/mol

= 0.03 g

Hence, percentage of the cyanide = 0.03 g/0.1183 g * 100

= 25.35%

8 0

3 years ago

Suppose of copper(II) acetate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of acetate

uranmaximum [27]

Answer:

0.0714 M for the given variables

Explanation:

The question is missing some data, but one of the original questions regarding this problem provides the following data:

Mass of copper(II) acetate: $m_{(AcO)_2Cu} = 0.972 g$

Volume of the sodium chromate solution: $V_{Na_2CrO_4} = 150.0 mL$

Molarity of the sodium chromate solution: $c_{Na_2CrO_4} = 0.0400 M$

Now, when copper(II) acetate reacts with sodium chromate, an insoluble copper(II) chromate is formed:

$(CH_3COO)_2Cu (aq) + Na_2CrO_4 (aq)\rightarrow 2 CH_3COONa (aq) + CuCrO_4 (s)$

Find moles of each reactant. or copper(II) acetate, divide its mass by the molar mass:

$n_{(AcO)_2Cu} = \frac{0.972 g}{181.63 g/mol} = 0.0053515 mol$

Moles of the sodium chromate solution would be found by multiplying its volume by molarity:

$n_{Na_2CrO_4} = 0.0400 M\cdot 0.1500 L = 0.00600 mol$

Find the limiting reactant. Notice that stoichiometry of this reaction is 1 : 1, so we can compare moles directly. Moles of copper(II) acetate are lower than moles of sodium chromate, so copper(II) acetate is our limiting reactant.

Write the net ionic equation for this reaction:

$Cu^{2+} (aq) + CrO_4^{2-} (aq)\rightarrow CuCrO_4 (s)$

Notice that acetate is the ion spectator. This means it doesn't react, its moles throughout reaction stay the same. We started with:

$n_{(AcO)_2Cu} = 0.0053515 mol$

According to stoichiometry, 1 unit of copper(II) acetate has 2 units of acetate, so moles of acetate are equal to:

$n_{AcO^-} = 2\cdot 0.0053515 mol = 0.010703 mol$

The total volume of this solution doesn't change, so dividing moles of acetate by this volume will yield the molarity of acetate:

$c_{AcO^-} = \frac{0.010703 mol}{0.1500 L} = 0.0714 M$

8 0

3 years ago

The air you breathe is made of 21% oxygen and 78% nitrogen. The oxygen in air is the _____ and the nitrogen is the _____.

Flura [38]

Here are the answers that would best complete the given statement above. The air you breathe is made of 21% oxygen and 78% nitrogen. The oxygen in air is the SOLUTE and the nitrogen is the SOLVENT. In a given mixture, the reagent having the highest percentage of amount is always the solvent and the one or ones having the lower percentages are the solutes. Hope this answer helps.

6 0

3 years ago