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Jan 28 2010 07:34am
Formel : MgCl2 + 2AgNo3 ---> 2AgCl + Mg2 + (No3)2


you have 5 grams MgCl and 10 grams AgNo how many grams Agcl will it make ?




I will offer a price for the persone to solve this question, I want the right answere with the way you calculated it ;)


GL
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Jan 28 2010 01:21pm
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Jan 28 2010 02:23pm
Quote (alpha_hannen @ Jan 28 2010 01:34pm)
Formel :MgCl2 + 2AgNo3 ---> 2AgCl + Mg2 + (No3)2

you have 5 grams MgCl and 10 grams AgNo how many grams Agcl will it make ?


In order to solve this problem, you need to identify the limiting reagent.

In this case, despite there being more Silver Nitrate being used in the reaction, it is the limiting reagent. The limiting reagent is the substance that produces the least amount of product using the balanced equation, using 5 grams of MgCl2 a total of 15.05g of AgCl were produced, yet using only 10 grams of AgNO3 produced a total of 8.43g of AgNO3. Thus, Silver Nitrate (AgNO3 is the limiting reagent)

Amount of AgCl produced from 5 grams of MgCl2:

5g MgCl2 x (1mol MgCl2/95.211g) x (2 mol AgCl/1 mol MgCl2) x (143.32g/1mol AgCl) = 15.05g AgCl

Amount of AgCl produced from 10 grams of AgNO3:

10g AgNO3 x (1mol AgNO3/169.88g) x (2mol AgCl/2mol AgNO3) x (143.32g/1mol AgCl) = 8.43g AgCl

Since 10 g of Silver Nitrate produces the least amount of product AgCl, it is the limiting reagent and therefore only 8.43g AgCl are actually produced by the chemical reaction.

This should help.
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Jan 28 2010 02:26pm
Quote (Jazz_Thing @ Jan 28 2010 08:23pm)
In order to solve this problem, you need to identify the limiting reagent.

In this case, despite there being more Silver Nitrate being used in the reaction, it is the limiting reagent. The limiting reagent is the substance that produces the least amount of product using the balanced equation, using 5 grams of MgCl2 a total of 15.05g of AgCl were produced, yet using only 10 grams of AgNO3 produced a total of 8.43g of AgNO3. Thus, Silver Nitrate (AgNO3 is the limiting reagent)

Amount of AgCl produced from 5 grams of MgCl2:

5g MgCl2 x (1mol MgCl2/95.211g) x (2 mol AgCl/1 mol MgCl2) x (143.32g/1mol AgCl) = 15.05g AgCl

Amount of AgCl produced from 10 grams of AgNO3:

10g AgNO3 x (1mol AgNO3/169.88g) x (2mol AgCl/2mol AgNO3) x (143.32g/1mol AgCl) = 8.43g AgCl

Since 10 g of Silver Nitrate produces the least amount of product AgCl, it is the limiting reagent and therefore only 8.43g AgCl are actually produced by the chemical reaction.

This should help.


ty ty ty
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Jan 29 2010 08:22am
What the hell is that?
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Jan 29 2010 09:06am
Quote (xVile @ Jan 29 2010 02:22pm)
What the hell is that?


thats the answer to the question :) he helped me calculate the grams of agcl that were created :)
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