First, we write the balanced equation for this reaction:
2KI + Pb(NO₃)₂ → 2KNO₃ + PbI₂
From this equation, we see that there are 2 moles of potassium iodide required for each mole of lead (II) nitrate. Moreover, we may use the formula:
Moles = volume (in L) * molarity
We find the molar relation ship for KI : Pb(NO₃)₂ to be 2 : 1. So:
M₁V₁ = 2M₂V₂
V₁ = 2M₂V₂/M₁
V₁ = 2 * 0.112 * 0.155 / 0.2
V₁ = 0.1736 L
The volume required is 173.6 mL
Answer:
The answer is D part of your lab notebook used for writing vocabulary
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Answer:
Mass = 255 g
Explanation:
Given data:
Number of moles of nitrogen = 7.5 mol
Mass of ammonia formed = ?
Solution:
Chemical equation:
3H₂ + N₂ → 2NH₃
Now we will compare the moles of nitrogen and ammonia.
N₂ : NH₃
1 : 2
7.5 : 2/1×7.5 = 15
Mass of ammonia:
Mass = number of moles × molar mass
Mass = 15 mol × 17 g/mol
Mass = 255 g
