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### Topic: How to find the maximum number of grams produced and grams left of a solution?  (Read 8231 times)

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#### remeday86

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##### How to find the maximum number of grams produced and grams left of a solution?
« on: October 27, 2008, 12:16:40 AM »

1)

Mass of aluminum metal used ---> 0.469 g
Mass of alum obtained --> 4.51 g
Molar mass of Aluminum used ---> 0.017 moles Al
Molar mass of alum (potassium aluminum sulfate) , KAl(SO4)2 * 12H2O ---> 474.41 g/mole

Noting that one aluminum atom is in each formula unit of alum, we know that each mole of aluminum will give rise to one mole of alum. Based on the number of moles of aluminum you used, calculate the maximum number of grams of alum that you produce, assuming that aluminum is the limiting reagent: ______ g alum.

This is what I got --> 8.06 g alum

0.017 moles Al x (1 mole Alum / 1 mole Al) x (474.4 g Alum/ 1 mole Alum) = 8.06 g alum

#### Borek

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##### Re: How to find the maximum number of grams produced and grams left of a solution?
« Reply #1 on: October 27, 2008, 04:50:34 AM »
Approach is correct, but you have rounded down intermediate value and that lead to a huge error in the final result.
« Last Edit: October 27, 2008, 05:23:36 AM by Borek »
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#### remeday86

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##### Re: How to find the maximum number of grams produced and grams left of a solutio
« Reply #2 on: October 27, 2008, 05:29:04 AM »
Approach is correct, but you have rounded down intermediate value and that lead to a huge error in the final result.

I am not sure what you mean that i have "rounded down intermediate value"?

#### Borek

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##### Re: How to find the maximum number of grams produced and grams left of a solution?
« Reply #3 on: October 27, 2008, 06:47:21 AM »
You calculated final mass using 0.017 moles, when the real value is something like 0.01738 (0.469/26.98)

In general, when calculating, you should give the final result with correct number of significant digits, but you should make calculations in full precision. Rounding down something in between is asking for errors.
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