# Chapter 4 Stoichiometry: Quantitative Information about Chemical Reactions - Study Questions - Page 179i: 107

#### Work Step by Step

Number of moles of $I_2$: $0.246\ M\times 40.21\ mL/1000 \ mL/L=0.00989\ mol$ From stoichiometry: $0.00989\ mol\times 2/1=0.01978\ mol$ thiosulfate. Mass of sodium thiosulfate: $0.01978\ mmol\times 158.10\ g/mol=3.127\ g$ Mass percentage: $3.127/3.232\times 100\%=96.76\%$

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