50.0kg of nitrogen and 10.0kg 0f hydrogen are mixed to produce ammonia . calculate the ammonia formed, identify the limiting reagent in the production of ammonia in this situation

Nitrogen is limiting reactant and 30.4kg of ammonia can be formed

Explanation:

Nitrogen, N₂, reacts with hydrogen, H₂ to produce ammonia, NH₃, as follows:

2N₂ + 3H₂ → 2NH₃

Where 2 moles nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia

To solve this question we must find the moles of each reactant in order to find limiting reactant. With limiting reactant we can find the moles and the mass of ammonia formed as follows:

Moles N2 -Molar mass: 28g/mol-

50000g * (1mol / 28g) = 1786 moles N2

Moles H2 -Molar mass: 2g/mol-

10000g * (1mol / 2g) = 5000 moles H2

For a complete reaction of 5000 moles H2 are needed:

5000 mol H2 * (2mol N2 / 3mol H2) = 3333 moles N2. As there are just 1786 moles, Nitrogen is limiting reactant

The moles of ammonia that can be produced are 1786 moles because 2mol N2 = 2moles NH3.

The mass of ammonia -Molar mass NH3: 17g/mol- is:

1786 moles NH3 * (17g / mol) = 30362g =

30.4kg of ammonia can be formed

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2