help with question

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sandraabdelmalak1D
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help with question

Postby sandraabdelmalak1D » Wed Jul 05, 2017 9:09 pm

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)

Okay so I get to the point where I see that C6H9Cl3 is the limiting reagent but then I don't know what to do. Can someone walk me through the rest of this problem?

Chels Zh 1D
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Re: help with question

Postby Chels Zh 1D » Wed Jul 05, 2017 9:33 pm

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Ryan Sydney Beyer 2B
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Re: help with question

Postby Ryan Sydney Beyer 2B » Tue Oct 03, 2017 11:58 am

To find the limiting reagent you need to convert both grams of C6H9Cl3 and AgNO3 to moles. So you take the 0.750g C6H9Cl3 and divide it by its molar mass of 187.50 g/mol and you'll have 0.004 mol C6H9Cl3. Do the same for AgNO3 which will be 1000. g AgNO3 divided by its molar mass of 169.88 g/mol and you'll have 5.887 mol AgNO3.

0.750g C6H9Cl3 * (1 mol / 187.50 g) = 0.004 mol C6H9Cl3

1000. g AgNO3 * (1 mol / 169.88 g) = 5.887 mol AgNO3

You need to compare these two mole quantities using the balanced equation :: C6H9Cl3 + 3AgNO3 ---> 3AgCl + C6H9(NO3)3

It can be seen by this balanced equation that 1 mol of C6H9Cl3 needs 3 mol of AgNO3 to react evenly. We previously determined that we have 0.004 mol of C6H9Cl3 which would mean that we would need 3 times as many moles of AgNO3 to react according to the equation. So we take the 0.004 mol and multiply by 3 which will tell us that we need at least 0.012 mol of AgNO3 to react evenly. We previously determined that we have 5.887 mol AgNO3 so therefore the C6H9Cl3 becomes the limiting reagent.

0.004 mol C6H9Cl3 * 3 = 0.012 mol ; 0.012 mol < 5.887 mol AgNO3


To find the mass of AgCl produced, we have to take the moles of C6H9Cl3 and transition it through the stoichiometric coefficients into moles of AgCl and eventually grams of AgCl. We have 0.004 mol C6H9Cl3 and to turn it into mol of AgCl we multiply by 3 because we can tell from the balanced equation that there are 3 mol of AgCl for every 1 mol of C6H9Cl3. We now have 0.012 mol AgCl and to turn this into grams we multiply by the molar mass of 143.32 g/mol. This leaves us with 1.72g AgCl.

0.004 mol C6H9Cl3 * (3 mol AgCl / 1 mol C6H9Cl3) * ( 143.32g AgCl / 1 mol AgCl) = 1.72 g AgCl


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