Consider the following reaction

Chemistry

1 answer:

aliina [53]3 years ago

7 0

Answer:

d, 40 dm3.

Explanation:

According to Avogadro's law, the mole ratio of chemicals in a reaction is equal to the ratio of volumes of chemicals reacted (for gas).

From the equation, the mole ratio of N2 : H2 : NH3 = 1 : 3 : 2, meaning 1 mole of N2 reacts completely with 3 moles of H2 to give 2 moles of NH3, the ratio of volume required is also equal to 1 : 3 : 2.

Considering both N2 and H2 have 30dm3 of volume, but 1 mole of N2 reacts completely with 3 moles of H2, so we can see H2 is limiting while N2 is in excess. Using the ratio, we can deduce that 10dm3 equals to 1 in ratio (because 3 moles ratio = 30dm3).

With that being said, all H2 has reacted, meaning there's no volume of H2 left. 2 moles of NH3 is produced, meaning the volume of NH3 produced = 10 x 2 = 20 dm3. (using the ratio again)

1 mole of N2 has reacted, meaning from the 30dm3, only 10 dm3 has reacted. This also indicate that 20 dm3 of N2 has not been reacted.

So at the end, the mixture contains 20dm3 of NH3, and 20 dm3 of unreacted N2. Hence, the answer is d, 40 dm3.

You might be interested in

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:

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