I need help with this one problem
I don't just need the answer I help to figure out how to solve it step by step it would help me soooo much
5.0 grams magnesium metal is dropped into a solution of silver nitrate, forming silver metal and magnesium nitrate. How many grams of magnesium nitrate is formed?
I got this for my balanced equation: Mg+2 AgNO3---> 2Ag+Mg(NO3)2 is this a good place to start and if it is, where do I go from here?
Your help is apperciated!
2007-12-28 17:53:43 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
does anyone one know of some generic formula for figuring this out?
2007-12-28 18:42:46 · update #1
well..ur first step is absolutely correct...
well..initially wat u do is...find out the "molecular weights" of magnesium and magnesium nitrate!...magnesium haz a molecular weight of 24 gms...and magnesium nitrate haz a weight of...148 gms...that is the initial case...i.e.from the balanced equation...
noe start comparing...initially wen the weight of Mg iz 24 gms..dat time the weight of Mg(NO3)2 is 148 gms..noe wen the given weight iz 5 gms...then wat is d weight of magnesium nitrate??..
24 ------148
5 --------?
so the answer comes out to b something around 30.833 gms...
i hope i m rite!..
-spooky
2007-12-28 18:12:39 · answer #1 · answered by ¿Spooky - Version 0.2? 3 · 0⤊ 0⤋
correct Mg +2AgNO3---> 2Ag +Mg (NO3)2
you see that 24g Mg corresponds to
24+2*(14+3*16)=148g Mg (NO3)2
1g corresponds to 148/24
and 5g 5*148/24=30.83g
Answer 30.83g of magnesium nitrate
2007-12-28 18:11:32 · answer #2 · answered by maussy 7 · 0⤊ 1⤋