Question 1
1 pts
What is the result when 12 grams of H2 and 28 grams of Nz react to
completion at STP.
3H2(g) + N2 --> 2 NH3
44 liters of NH3 are produced and 66 liters of H2 remain
22 liters of NH3 are produced and no liters of H2 remain
44 liters of NH3 are produced and no reactants remain
22 liters of NH3 are produced and 66 liters of N2 remain

Question:

a. 44 liters of NH3 are produced and 66 liters of H2 remain

b. 22 liters of NH3 are produced and no liters of H2 remain

c. 44 liters of NH3 are produced and no reactants remain

d. 22 liters of NH3 are produced and 66 liters of N2 remain

The most correct option is;

a. 44 liters of NH₃ are produced and 66 liters of H₂ remain

Explanation:

We are given the equation of the reaction as follows;

3H₂ (g) + N₂ (g) → 2NH₃

Therefore, 3 moles of H₂ reacts with 1 mole of N₂ to produce 2 moles of NH₃

Molar mass of H₂ = 2.01588 g/mol

Molar mass of N₂ = 28.0134 g/mol

Molar mass of NH₃ = 17.031 g/mol

Therefore

Number of moles of H₂ =

∴ Number of moles of H₂ ≈ 6 moles

Number of moles of N₂ =

∴ Number of moles of N₂ ≈ 1 mole

Since 3 moles of H₂ reacts with 1 mole of N₂ to produce 2 moles of NH₃,  we have an excess of approximately 3 moles of H₂ with 2 moles of NH₃ produced

From the universal gas equation we have;

Where:

For NH₃

n = Number of moles  = 2

P = Pressure at STP = 1 atm

V = Volume = Required

T = Temperature at STP = 273.15 K

R = Universal Gas Constant =0.08205  L·atm/(mol·K)

Plugging the values, we have;

For the excess H₂

n = Number of moles  = 3

P = Pressure at STP = 1 atm

V = Volume = Required

T = Temperature at STP = 273.15 K

R = Universal Gas Constant =0.08205  L·atm/(mol·K)

Plugging the values, we have;

Therefore, the most suitable option is 44 liters of NH₃ are produced and 66 liters of H₂ remain.

c. orbital shape i hope this is correct