Question

what is the mass of silver in 3.4 g AgNo3

Answer 1

I fast do the meta work:
\[\Large n(AgNO_{3})=\frac{ m(AgNO_{3}) }{ M(AgNO_{3}) }\]
\[\Large n(AgNO_{3})=n(Ag (s))\]
\[\Large n(Ag)=\frac{ m(Ag) }{ M(Ag) }\]
Put substitute the equations and calculate the mass of silver.

Sorry for the quick and perhaps a bit unhelping answer.

Answer 2

molecular mass of Agno3 contain=molecular mass of Ag
i.e 170 g of Agno3 = 108 g of Ag
3.4 g of Agno3 = 108 / 170 * 3.4 = 2.16 g of Ag.

Answer 3

that is unity method... nd easy method

Answer 4

Well the setup I've done is the similar unity method or what ever that is... set it all together you get
\[\Large \frac{ m(Ag) }{ M(Ag) }=\frac{ m(AgNO_{3}) }{ M(AgNO_{3}) }\]
Rearange
\[\Large m(Ag)=\frac{ M(Ag) }{ M(AgNO_{3}) } \times m(AgNO_{3})\]
Exactly what you wrote.

Answer 5

What I could fear is that the unity method could be used incorrectly if they got a question where it wasn't a 1:1 ratio. For example if they were asked to calculate the mass contribution of oxygen.

Answer 6

I agree bro....

Answer 7

then we can also get for Oygen.
i.e 170 g of AgNO3 = 3 O
170 g of AgNO3 = 3 *16 = 48g
3.4 gram of AgNo3 = 48 / 170 * 3.4 = 0.96 g of Oxygen...
Plzzz check it and tell me is it right???? Mr @Frostbite

Answer 8

Looks correct! :) It was just the 3*16 I could be afraid others would forget. I don't doubt you can do the question easily.

Answer 9

ok thanku bro :)