High School

Thank you for visiting If 1 70 g of Pb NOâ ƒ â are reacted with 3 40 g of KI how many grams of PbIâ will be produced. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!

If 1.70 g of Pb(NO₃)₂ are reacted with 3.40 g of KI, how many grams of PbI₂ will be produced?

Answer :

In this casw, the mass of PbIâ‚‚ produced will be approximately 3.43g.

Given:

Mass of Pb(NO₃)₂ = 1.70 g

Mass of KI = 3.40 g

To find:

Mass of PbIâ‚‚ produced

Step-by-Step Solution:

Calculate the molar mass of each compound:

Pb(NO₃)₂:

  • Pb: 207.2 g/mol
  • N: 14.01 g/mol
  • O: 16.00 g/mol

Molar mass = 207.2 + 2(14.01 + 3(16.00)) = 331.2 g/mol

KI:

K: 39.10 g/mol

I: 126.90 g/mol

Molar mass = 39.10 + 126.90 = 166.00 g/mol

Determine the moles of each reactant:

Moles of Pb(NO₃)â‚‚ = Mass / Molar mass = 1.70 g / 331.2 g/mol ≈ 0.00513 mol

Moles of KI = Mass / Molar mass = 3.40 g / 166.00 g/mol ≈ 0.0205 mol

Identify the limiting reactant by comparing the mole ratios from the balanced chemical equation:

The balanced chemical equation for the reaction is:

Pb(NO₃)â‚‚ + 2Kl → Pblâ‚‚ + 2KNO₃

From the equation, it is a 1 : 2 ratio between Pb(NO₃)₂ and KI.

Since there are fewer moles of Pb(NO₃)₂ compared to KI, Pb(NO₃)₂ is the limiting reactant.

Calculate the theoretical yield of PbIâ‚‚ using the limiting reactant:

Moles of PbI₂ produced will be half that of moles of limiting reactant (Pb(NO₃)₂).

Moles of PbIâ‚‚ = Moles of Pb(NO₃)â‚‚ / 1/2 ≈ 0.00513mol/ (1/2)

Moles of PbIâ‚‚ ≈ 0.01026 mol

Convert moles of PbIâ‚‚ to grams using its molar mass:

Molar mass of PbIâ‚‚:

Pb: 207.2 g/mol

I: 126.90 g/mol

Molar mass = 207.2 + 126.90 = 334.10 g/mol

Thank you for reading the article If 1 70 g of Pb NOâ ƒ â are reacted with 3 40 g of KI how many grams of PbIâ will be produced. We hope the information provided is useful and helps you understand this topic better. Feel free to explore more helpful content on our website!

Rewritten by : Jeany

Other Questions