# Chapter 3 - Mass Relationships in Chemical Reactions - Questions & Problems - Page 109: 3.65

3.6 mol $CO_{2}$

#### Work Step by Step

$2CO(g)+0_{2}(g)$ -> $2CO_{2}(g)$ mol of $CO_{2}$= $\frac{(3.6 mol CO)\times(2 mol CO_{2})}{(2 mol CO)}$ = 3.6 mol $CO_{2}$

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