how many moles of NaBr are produced from reacting 20.05 moles of AuBr3 with excess NaCN?
AuBr3 +NaCN -> Au(CN)3 + NaBr
To determine the number of moles of NaBr produced, we need to use the balanced chemical equation and the stoichiometry of the reaction.
First, let's identify the stoichiometric ratio between AuBr3 and NaBr from the balanced equation:
AuBr3 + 3 NaCN → Au(CN)3 + 3 NaBr
From the equation above, we can see that for every 1 mole of AuBr3, 3 moles of NaBr are produced.
Given that we have 20.05 moles of AuBr3, we can calculate the number of moles of NaBr produced by multiplying the number of moles of AuBr3 by the stoichiometric ratio:
20.05 moles AuBr3 × (3 moles NaBr / 1 mole AuBr3) = 20.05 moles AuBr3 × 3 = 60.15 moles NaBr.
Therefore, 60.15 moles of NaBr are produced from reacting 20.05 moles of AuBr3 with excess NaCN.