how many moles of NaBr are produced from reacting 20.05 moles of AuBr3 with excess NaCN?
AuBr3 +NaCN -> Au(CN)3 + NaBr
To find out how many moles of NaBr are produced, we need to use the balanced chemical equation and the stoichiometry of the reaction.
The balanced chemical equation you provided is:
AuBr3 + 3NaCN -> Au(CN)3 + 3NaBr
This equation tells us that for every 1 mole of AuBr3, 3 moles of NaBr are produced.
Given that you have 20.05 moles of AuBr3, we can set up a proportion to find the number of moles of NaBr produced:
1 mole AuBr3 / 3 moles NaBr = 20.05 moles AuBr3 / x moles NaBr
Cross-multiplying the proportion, we get:
1 * x = 3 * 20.05
x = 60.15 moles NaBr
Therefore, 60.15 moles of NaBr are produced from reacting 20.05 moles of AuBr3 with excess NaCN.