Question

): Given the following reaction: Na2S2O3 + AgBr NaBr + Na3[Ag(S2O3)2]
a. How many moles of Na2S2O3 are needed to react completely with 42.7 g of AgBr?
b. What is the mass of NaBr that will be produced from 42.7 g of AgBr?

Answer 1

Explanation:

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Answer 2

Answer:

0.455 moles

Explanation:

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