Answered: In one experiment, 12.0 g of PCl5 was…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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In one experiment, 12.0 g of PCl5 was slowly added to 15.0 g of H2O according to the following balanced chemical equation:

Pcl5(s)+4H2O(l) ---> H3PO4 (aq) + 5HCl (g)

The molar masses for each compound in the equation are as follows;

PCl5: 208.224 g/mol

H2O: 18.015 g/mol

H3PO4: 97.994 g/mol

HCl; 36.45 g/mol

What is the limiting reagent in this scenario?

The answer given to me was:

From, the above equation, it is observed that 1 mole of PCl5 is reacted with 4 moles of H2O.

Then, 0.058 moles of PCl5 are reacted with

0.058 mol/1 mol × 4 mol= 0.232 moles of H2O.

But the actual number of moles of water available is 0.833 mol. So, in this chemical reaction, water is available in excess amount and PCl5 is present in a lesser amount. So, all the PCl5 is consumed in this reaction, as it is present in less amount compared to the other reactant. And, hence, the limiting reagent is PCl5. Now, the excess amount of water is = (0.833 - 0.232) mol = 0.601 mol.

Could you elaborate on the answer? What is canceling out that dimensional analysis conversion factor for 4x0.058mol /1 mol? Could you label them with units? Example attached.

Scenario #1
How many grams of silver chloride (AgCl) are produced from the reaction between
100.0 g of silver nitrate (AgNO,) and 100.0 g barium chloride (BaCl₂) ??
2AgNO3(s) + BaCl₂ (s)
moles
liters
100.0 g
100 g BaCl2 x
100.0 g
Determine the number of moles of product that can be created from 100 g of AgNO₂:
100.0 g-AgNO3 x
1 mole AgNO3
169.87 moles AgNO3
What is the
limiting reactant?
AgNO3
Ba(NO3)2 (5) + 2AgCl (s)
0.5887 moles
1 moles BaCl2
Determine the number of moles of product that can be created from 100 g of BaCl,:
2 moles AgC!
X
1mol Bach
208.23 g BaCl2
X
2 moles AgCl
2 moles AgNO3
0.5887 moles AgCl X 143.2 g AgCl
1 mole AgCl
MOLAR MASSES IF NEEDED:
169.87 g/mol
143,32 g/mol
213.341 g/mol
208.23 g/mol
0.58868 moles
- 200/333.74= 0.5887 moles AgCl
What is the maximum amount of
AgCl that can be formed?
AgNO.
AgCl
Ba(NO₂)₂
200.0/208.23 =
84.37 g AgCl
0.960476 moles AgCl
0.9606 molos
AgCl
0.5887 moles AgCl

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