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1. If 358 g of sodium undergoes a single-displacement reaction with excess gold(II) chloride, how many grams of elemental gol
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Answer #1

2 Na + AuCl2 -> 2 NaCl + Au

Molar mass of Na = 22.99 g/mol

mass of Na = 3.58*10^2 g

mol of Na = (mass)/(molar mass)

= 3.58*10^2/22.99

= 15.57 mol

Balanced chemical equation is:

2 Na + AuCl2 -> 2 NaCl + Au

According to balanced equation

mol of Au formed = (1/2)* moles of Na

= (1/2)*15.57

= 7.786 mol

Molar mass of Au = 197 g/mol

mass of Au = number of mol * molar mass

= 7.786*1.97*10^2

= 1.534*10^3 g

Answer: 1.53*10^3 g

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