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

water at room temperature is flowing through a pipe with pressure taps located 10 ft apart along one section of the pipe, which are connected to a mercury (s=13.6) manometer.