The Synthesis of Ammonia is done through Haber's Process:
N2 + 3H2 -> 2NH3
so 2 moles of NH3 requires 3 moles of H2 gas
so if X moles of NH3 requires 19.5 moles of H2 gas
then X=(2/3)*19.5=13
NH3 formed = 13 moles
4 moles
30 moles
The answer is 30 moles.
None, unless there is metallic potassium in the reaction mixture. Assuming excess potassium metal is present then 14 moles of KBr can be produced. 7BaBr2 + excess potassium -----> 14KBr + 7 Ba
11,2 moles carbon dioxide are obtained.
6 moles
X = 0.489 moles of AgCl produced
30 moles
4 moles
4 moles
30 moles
The answer is 30 moles.
None, unless there is metallic potassium in the reaction mixture. Assuming excess potassium metal is present then 14 moles of KBr can be produced. 7BaBr2 + excess potassium -----> 14KBr + 7 Ba
8 mol
11,2 moles carbon dioxide are obtained.
Equation. 2Al + 3Cl2 -> 2AlCl3 one to one again 0.440 moles Al (2 moles AlCl3/2 moles Al) = 0.440 moles AlCl3 produced
The reaction is:WO3 + 3 H2 = W + 3 H2OThree moles of hydrogen for one mole of wolfram.