Best Answer
1 mole of a substance contains 6.022x10^23 "units" of that substance.
So 0.187 mol of Na+ is 1.13x10^23 ions (6.022x10^23 x 0.187).
There are 137 atoms in this molecule. C55 + H72 = 127. 127 + Mg (one atom of magnesium = 128. 128 + N4 = 132. 132 + O5 = 137.
Answer:
3.15 × 10⁻⁶ mol H₂/L.s
1.05 × 10⁻⁶ mol N₂/L.s
Explanation:
Step 1: Write the balanced equation
2 NH₃ ⇒ 3 H₂ + N₂
Step 2: Calculate the rate of production of H₂
The molar ratio of NH₃ to H₂ is 2:3. Given the rate of decomposition of NH₃ is 2.10 × 10⁻⁶ mol/L.s, the rate of production of H₂ is:
2.10 × 10⁻⁶ mol NH₃/L.s × 3 mol H₂/2 mol NH₃ = 3.15 × 10⁻⁶ mol H₂/L.s
Step 3: Calculate the rate of production of N₂
The molar ratio of NH₃ to N₂ is 2:1. Given the rate of decomposition of NH₃ is 2.10 × 10⁻⁶ mol/L.s, the rate of production of N₂ is:
2.10 × 10⁻⁶ mol NH₃/L.s × 1 mol N₂/2 mol NH₃ = 1.05 × 10⁻⁶ mol N₂/L.s
Reactant C is the limiting reactant in this scenario.
Explanation:
The reactant in the balanced chemical reaction which gives the smaller amount or moles of product is the limiting reagent.
Balanced chemical reaction is:
A + 2B + 3C → 2D + E
number of moles
A = 0.50 mole
B = 0.60 moles
C = 0.90 moles
Taking A as the reactant
1 mole of A reacted to form 2 moles of D
0.50 moles of A will produce
= 
thus 0.50 moles of A will produce 1 mole of D
Taking B as the reactant
2 moles of B reacted to form 2 moles of D
0.60 moles of B reacted to form x moles of D
= 
x = 2 moles of D is produced.
Taking C as the reactant:
3 moles of C reacted to form 2 moles of D
O.9 moles of C reacted to form x moles of D
= 
= 0.60 moles of D is formed.
Thus C is the limiting reagent in the given reaction as it produces smallest mass of product.