Question

A sample of 11.7 moles of silver bromide is reacted

with 8.6 moles of sodium thiosulfate to produce
sodium silver thiosulfate and sodium bromide. Using
the balanced equation below, predict which is the
limiting reactant and the maximum amount in moles
of sodium bromide that can be produced.

AgBr + 2Na2S2O3 → Na3Ag (S₂03)2 + NaBr

A. sodium thiosulfate, 4.30 moles
B. sodium thiosulfate, 8.60 moles
C. sodium bromide, 5.85 moles
D. sodium bromide, 11.7 moles

Answer 1

0.30 moles of sodium thiosulfate formula units are needed to make 0.10 moles of AgBr soluble.

Step-by-step explanation:

3Na2S2O3 + Ag Br→ NaBr + Nas [Ag(S2O3)3 According to reaction, 1 mole of silver bromide dissolves in 3 moles of Sodium thiosulfate. Then 0.10 moles of AgBr will dissolve with:3/1 x 0.10mol = 0.30mol - 0.30 moles of sodium thiosulfate formula units are needed to make 0.10 moles of AgBr soluble. Hope it will be