How much energy is lost through carrying out life functions?

If 100.0g of nitrogen is reacted with 100.0g of hydrogen, what is the theoretical yield of the reaction? What is the excess reac

Shkiper50 [21]

theoretical yield of ammonia (NH₃) = 121.38 g

The limiting reactant is nitrogen (N₂) and the excess reactant is hydrogen (H₂).

Explanation:

We have the following chemical reaction in which nitrogen react with hydrogen to produce ammonia:

N₂ + 3 H₂ → 2 NH₃

Now we need to calculate the number of moles of each reactant:

number of moles = mass / molar weight

number of moles of N₂ = 100 / 28 = 3.57 moles

number of moles of H₂ = 100 / 2 = 50 moles

We see from the chemical reaction that 3 moles of H₂ react with 1 mole of N₂ so 50 moles of H₂ react with 16.67 moles of N₂ which is way more than the available N₂ quantity of 3.57 moles, so the limiting reactant is nitrogen (N₂) and the excess reactant is hydrogen (H₂).

Knowing this we devise the following reasoning:

if 1 mole of N₂ produces 2 moles of NH₃

then 3.57 moles of N₂ produces X moles of NH₃

X = (3.57 × 2) / 1 = 7.14 moles of NH₃

mass = number of moles × molar weight

mass of NH₃ = 7.14 × 17 = 121.38 g (theoretical yield)

Learn more about:

limiting reactant

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7 0

3 years ago

When cyclopentane undergo free-radical substitution with bromine (Br2 /Heat) the product:

Artemon [7]

Explanation:

b. Bromo cyclopentane + HBr

4 0

3 years ago

Phosphorous pentoxide, P2O5(s), is produced from the reaction between pure oxygen and pure phosphorous (P, solid). What is the v

BartSMP [9]

Answer:

4190.22 L = 4.19 m³.

Explanation:

For the balanced reaction:

2P₂ + 5O₂ ⇄ 2P₂O₅.

It is clear that 2 mol of P₂ react with 5 mol of O₂ to produce 2 mol of P₂O₅.

Firstly, we need to calculate the no. of moles of 6.92 kilograms of P₂O₅ produced through the reaction:

no. of moles of P₂O₅ = mass/molar mass = (6920 g)/(283.88 g/mol) = 24.38 mol.

Now, we can find the no. of moles of O₂ is needed to produce the proposed amount of P₂O₅:

Using cross multiplication:

5 mol of O₂ is needed to produce → 2 mol of P₂O₅, from stichiometry.

??? mol of O₂ is needed to produce → 24.38 mol of P₂O₅.

∴ The no. of moles of O₂ needed = (5 mol)(24.38 mol)/(2 mol) = 60.95 mol.

Finally, we can get the volume of oxygen using the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm (P = 606.1 mm Hg/760 = 0.8 atm).

V is the volume of the gas in L (V = ??? L).

n is the no. of moles of the gas in mol (n = 60.95 mol).

R is the general gas constant (R = 0.0821 L.atm/mol.K),

T is the temperature of the gas in K (396.90°C + 273 = 669.9 K).

∴ V of oxygen needed = nRT/P = (60.95 mol)(0.0821 L.atm/mol.K)(669.9 K)/(0.8 atm) = 4190.22 L/1000 = 4.19 m³.

3 0

3 years ago

En que epoca aprendieron a trabaja el cobre y otros metales?

g100num [7]

Maybe if you put it in english i’ll answer.. nahh jp i’m fluent

6 0

2 years ago

Determine the electron geometry (eg) and molecular geometry (mg) of CO2.

eimsori [14]

Answer:

eg=linear, mg=linear

Explanation:

First of all, it must be stated that most triatomic molecules are either linear or bent. This depends on the electron geometry of the molecule and the number of bonding groups, multiple bonds and lone pairs present.

CO2 contains four regions of electron density and two bonding groups. For a specie containing two bonding groups, a linear molecular geometry is expected with an angle of 180°.

For a specie having two bonding groups and no lone pairs with multiple bonds, the expected electron geometry is also linear.

6 0

3 years ago