## Problem 10 on Limiting Reactant [ENDORSED]

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Xin Huang 3E
Posts: 40
Joined: Wed Sep 21, 2016 2:57 pm

### Problem 10 on Limiting Reactant

This is Problem #10 on Pre-Module Assessment:
According to the following equation, 0.750 g of C6H9Cl3 is mixed with 1.000 kg of AgNO3 in a flask of water. A white solid, AgCl, completely precipitates out. What is the mass of AgCl produced?
C6H9Cl3 + 3AgNO3 ---> AgCl + C6H9(NO3)3.
Molar Mass: C6H9Cl3 (187.50 g/mol), 3AgNO3 (169.88 g/mol), AgCl (143.32 g/mol)
I got the result of 0.57g AgCl (Calculation: it's obvious that C6H9Cl3 is the limiting reactant, so 0.750g/(187.50g/mol)=x/(143.32 g/mol) , and solve the x)but the answer choices are:
A. 1.00 g AgCl
B. 1.14 g AgCl
C. 1.50 g AgCl
D. 1.72 g AgCl
So obviously either I calculated it wrong or the answer choices are wrong...but I think there is higher possibility that I got it wrong. But what should be the correct process of calculation? Are there anyone having similar problem? (I did this assessment a couple of week ago...so don't mark me for late homework : )

Calvin Huang 3D
Posts: 16
Joined: Fri Jul 15, 2016 3:00 am
Been upvoted: 1 time

### Re: Problem 10 on Limiting Reactant

First thing you should check for is if your chemical equation is balanced and then continue on with your calculations.

Molly_McMillen_3J
Posts: 9
Joined: Wed Sep 21, 2016 2:57 pm

### Re: Problem 10 on Limiting Reactant  [ENDORSED]

As the previous poster said, double check that your equation is balanced because I got that the balanced equation is
C6H9Cl3 + 3 AgNO3 --> 3AgCl + C6H9(NO3)3
Then what I did next was the following:
0.750g C6H9Cl3 x 1mole/187.5g= 0.004 moles x (1mol C6H9Cl3/ 3 mole AgCl) x (143.32g AgCl/ 1 mole AgCl)= 1.72g AgCl

Xin Huang 3E
Posts: 40
Joined: Wed Sep 21, 2016 2:57 pm

### Re: Problem 10 on Limiting Reactant

Oops, my bad. Thank you for pointing out my flaws