How many moles of Na2S2O3 are needed to react completely with 42.7 g of AgBr?
Answers
0.455 moles is answer and explanation hope it helps please mark me as Assuming that the reaction you meant was:
Na2S2O3+AgBr→Na3(Ag(S2O3)2)+NaBr
If I'm wrong tell me in the comments.
Step One: Write balanced equation
The reaction above is not balanced. The balanced equation would be:
2Na2S2O3+AgBr→Na3(Ag(S2O3)2)+NaBr
Step Two: Find molar ratio between substances
Using the above we can see that the molar ratio of
Na2S2O3:AgBr
is equal to
2:1
(As the number in front of the substances represents the molar ratios)
Step Three: Find the number of moles of AgBr
Using the formula:
n=m/m
we can find out the number of moles of AgBr
n=42.7/108+80
n=0.2274
Step Four: Calculate the number of moles required
As mentioned before, the ratio between
Na2S2O3:AgBr
is equal to
2:1
Therefore, the number of moles of Na2S2O3 required is equal to:
0.2274×2=0.455