What is the volume of air required to completely burn 1 L of $C O$ ? Assume that all volumes are measured at the same temperature and pressure and also that air contains one-fifth by volume of $O_{2}$

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Solution

$C O + \frac{1}{2} O_{2} + N_{2} \to C O_{2} + N_{2}$
given {20% oxygen , 80% nitrogen}
But 1 mol CO Need = $\frac{1}{2} m o l o f O_{2}$
$1 l i t C O = \frac{1}{2} l i t o f O_{2}$
air required = $\frac{100}{20} \times$ oxygen Needed
$= \frac{100}{20} \times \frac{1}{2}$
$= 2.5 l i t$

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What is the volume of air required to completely burn 1 L of CO? Assume that all volumes are measured at the same temperature and pressure and also that air contains one-fifth by volume of O2

CO+12O2+N2→CO2+N2given {20% oxygen , 80% nitrogen}But 1 mol CO Need = 12molofO21litCO=12litofO2air required = 10020× oxygen Needed =10020×12 =2.5lit

What is the volume of air required to completely burn 1 L of $C O$ ? Assume that all volumes are measured at the same temperature and pressure and also that air contains one-fifth by volume of $O_{2}$

$C O + \frac{1}{2} O_{2} + N_{2} \to C O_{2} + N_{2}$
given {20% oxygen , 80% nitrogen}
But 1 mol CO Need = $\frac{1}{2} m o l o f O_{2}$
$1 l i t C O = \frac{1}{2} l i t o f O_{2}$
air required = $\frac{100}{20} \times$ oxygen Needed
$= \frac{100}{20} \times \frac{1}{2}$
$= 2.5 l i t$