Given molar mass of N2 is 28.02 g/mol
Molar mass of H2 is 2.02 g/mol
According to the stoichiometric equation 1mole of N2 reacts with 3moles of H2 to give 1mole of NH3
This means 1x28.02 gms of N2 reacts with 3x2.02 gms of H2
Given 5.42 gms of N2 and 5.42 gms of H2 are allowed to react
If 6.06 gms of H2 can react with 28.02 gms of N2 then
5.42 gms of H2 can react with
Mass of Nitrogen can be consumed = (5.42/6.06)×28.02 = 25.11 gms
Thus 6.06 gms of hydrogen can react with 25.11 gms of nitrogen. But the available nitrogen is only 5.42gms. As the availability of nitrogen become limited and still hydrogen is present, here nitrogen is the limiting reagent.
4. Consider the following chemical reaction, Na(s) + 3 Hale) - 2 NH,(g) [balanced] 15:42 g...
2 NH (g) + 3 CuO (s) à Ng) + 3 Cu(s) + 3H,0 (g) Molar Masses NH3: 17.03 Cuo: 79.55 N,: 28.02 H20: 18.02 A) Determine (with a clear supporting calculation) which is the limiting reagent. B) What mass of Cu(s) is formed? C) Only 40.3 g of Cu(s) are recovered. What is the % yield?
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