can u solve this g chem q (kaplan).

[Tina324]

---

Members do not see this ad.

If 5.8 g of Ag(NH3)2+ yields 1.4g of ammonia, how many moles of silver are produced?

MM of Ag(NH3)2+ =141.9 g/mol
MM of Ag= 107.9 g/mol

How would u guys go about solving this problem?

the answer is..........



.041

 

[nixon13]

this is a simple problem...so first write out the rxn equation.

so Ag(NH3)2+ --> Ag + NH3 all in 1:1:1 mole ratio as balanced.

So you have

5.8g Ag(NH3)2+ X (1 molAg(NH3)2+/141.9g Ag(NH3)2+)= mol Ag(NH3)2+

So you take that answer and use the mole ratios to see how much Ag will be produced.

So since its all 1:1:1 mole ratios you only have to do 5.8/141.9= .041g

 

[Optimism Smiles]

If 5.8 g of Ag(NH3)2+ yields 1.4g of ammonia, how many moles of silver are produced?

MM of Ag(NH3)2+ =141.9 g/mol
MM of Ag= 107.9 g/mol

How would u guys go about solving this problem?

the answer is..........



.041

This problem is really straightforward. You start with 5.8g Ag(NH3)2 and multiply by 1mol/141.9g Ag(NH3)2 to find the # moles of Ag(NH3)2 (which is .041 mol). Since there is 1mol Ag2+ produced per 1mol Ag(NH3)2 reacted (set up looks like 0.41 mol Ag(NH3)2 x 1 mol Ag2+/1 mol Ag(NH3)2 ), the answer is 0.041 mol silver.

A second way to do this works equally well but with the same principles. The balanced equation is:
Ag(NH3)2 -> 2NH3 + Ag2+

Convert the 1.4g NH3 to moles (1.4 g NH3 x 1mol/17g NH3) and you get .082 mol NH3. Since there are 1 mol Ag2+ produced per 2 mol NH3 produced, the answer is also .041 mol. (The setup is .082 mol NH3 x 1 mol Ag2+/2 mol NH3 = .041 mol Ag2+).

Hope this helps! 🙂

 

[Tina324]

thanks guys =)