# Chapter 4 Stoichiometry: Quantitative Information about Chemical Reactions - Study Questions - Page 179h: 95

Sn

#### Work Step by Step

Number of moles of $O$: $(0.452-0.356)\ g\div 15.9994\ g/mol= 6.00\ mmol$, $3.00\ mmol$ of M Atomic weight of M: $0.356\ g\div 0.003\ mol= 118.7\ g/mol$, closest to Sn

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