Question

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Consider the following balanced equation for the following reaction: 3H2SO4(aq) + 2Fe(s) + 3H2(g) + Fe2(SO4)3(aq) Determine the amount of H2(9) formed in the reaction if the percent yield of H2(g) is 65.0% and the theoretical yield of H2(g) is 39.0 grams. ООООО 15.2 grams 30.5 grams 25.4 grams 60.0 grams 43.2 grams N

Answer 1

Answer -

Given,

Theoretical Yield of H2 (g) = 39.0 g

Percentage Yield = 65.0 %

Actual Yield of H2 (g) = ?

We know that,

Percentage Yield = (Actual Yield/ Theoretical Yield) * 100

65.0 = (Actual Yield/ 39.0 g) * 100

0.65 = (Actual Yield/ 39.0 g)

So, Actual Yield = 0.65 * 39.0 g

Actual Yield = 25.35 g

So, Actual Yield of H2 (g) is 25.4 g [ANSWER]

Answer -

Given,

Percentage Yield of NaNO3 (aq) = 90.0 %

Actual Yield (moles) of NaNO3 (aq) = 0.877 mol

Molar Mass of NaNO3 (aq) = 85 g/mol

Theoretical Yield (grams) of NaNO3 (aq) = ?

We know that,

Percentage Yield = (Actual Yield/ Theoretical Yield) * 100

90.0 = (0.877 mol / Theoretical Yield) * 100

0.9 = (0.877 mol / Theoretical Yield)

Theoretical Yield = (0.877 mol / 0.9)

Theoretical Yield = 0.974 mol

Now,

Moles = mass / Molar Mass

So, Mass = Moles * Molar mass

Theoretical Yield of NaNO3 (aq) in g = 0.974 mol * 85 g/mol

Theoretical Yield of NaNO3 (aq) in g = 82.79 g

So, Theoretical Yield of NaNO3 (aq) is 82.8 g [ANSWER]

Answer -

Given,

mass of Urea = 177.5 kg or 177500 g [1 kg = 1000g]

Molar Mass of Urea = 60.06 g/mol

Percentage Yield of Over all reaction = 79.0 %

Molar Mass of Melamine = 126.12 g/mol

Mass of Melamine produced = ?

We know that,

Moles = Mass / Molar Mass

So, Moles of Urea = 177500 g / 60.06 g/mol

Moles of Urea = 2955.37 mol

Now,

CO(NH2)2 (l) $\rightarrow$ HNCO (l) + NH3 (g) -----1

6 HNCO (l) $\rightarrow$ C3N3(NH2)3 (l) + 3 CO2 ----2

Multiply 1 reaction by 6,

6 CO(NH2)2 (l) $\rightarrow$ 6 HNCO (l) + 6 NH3 (g) ---3

Now,

Insert reaction 2 in 3,

6 CO(NH2)2 (l) $\rightarrow$ C3N3(NH2)3 (l) + 3 CO2 + 6 NH3 (g)

Using Stiochiometry, It can be analyzed that for 6 mole of CO(NH2)2, 1 mole of melamine is produced.

i.e.

Moles of Melamine produced = (1/6)* Moles of Urea

Moles of Melamine produced = (1/6)* 2955.37 mol

Moles of Melamine produced = 492.562 mol

Also,

The percentage Yield = 79.0 %

So,

Actual moles of Melamine produced = 79.0% * 492.562 mol

Actual moles of Melamine produced = 0.79 * 492.562 mol

Actual moles of Melamine produced = 389.12 mol

Now,

Moles = mass / Molar Mass

Mass = Moles * Molar Mass

So, Mass of Melamine produced = 389.12 mol * 126.12 g/mol

Mass of Melamine produced = 49075.81 g

Now,

1 g = 0.001 kg

So, 49075.81 g = 49.1 kg

Mass of Melamine produced = 49.1 kg [ANSWER]