How many moles of $C O_{2}$ $CO_2$ (g) are in a 5.6 L sample of $C O_{2}$ $CO_2$ measured at STP?

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How many moles of $C O_{2}$ $CO_2$ (g) are in a 5.6 L sample of $C O_{2}$ $CO_2$ measured at STP?

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Answer 1 · 2016-01-30T17:21:02

$moles = 0.25$ $"moles"=0.25$

Explanation:

Using the ideal gas equation, we can solve for the number of moles:

$P V = n R T$ $PV=nRT$

where:
$P =$ $P=$ pressure
$V =$ $V=$ volume
$n =$ $n=$ moles
$R =$ $R=$ universal constant $(8.314 \frac{k P a \cdot L}{m o l \cdot K})$ $(8.314(kPa*L)/(mol*K))$
$T =$ $T=$ temperature (Kelvin)

Recall that at STP conditions:

$P = 101.325$ $P=101.325$ $k P a$ $kPa$
$T = 273.15$ $T=273.15$ $K$ $K$

To solve for the number of moles of carbon dioxide gas, substitute your known values into the ideal gas equation:

$P V = n R T$ $PV=nRT$

$n = \frac{P V}{R T}$ $n=(PV)/(RT)$

$n = \frac{(101.325 k P a) (5.6 L)}{(8.314 \frac{k P a \cdot L}{m o l \cdot K}) (273.15 K)}$ $n=((101.325kPa)(5.6L))/((8.314(kPa*L)/(mol*K))(273.15K))$

$n=((101.325color(red)cancelcolor(black)(kPa))(5.6color(orange)cancelcolor(black)(L)))/((8.314(color(red)cancelcolor(black)(kPa)*color(orange)cancelcolor(black)(L))/(mol*color(green)cancelcolor(black)(K)))(273.15color(green)cancelcolor(black)(K)))$

$n=0.2498580892$ $mol$

$n=0.25$ $mol$ (rounded to $2$ significant figures)

$:.$ , there are $0.25$ $mol$ in $5.6$ $L$ of $CO_(2(g))$ at STP conditions.

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How many moles of $C O_{2}$ $CO_2$ (g) are in a 5.6 L sample of $C O_{2}$ $CO_2$ measured at STP?

1 Answer

Explanation:

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How many moles of $C O_{2}$ $CO_2$ (g) are in a 5.6 L sample of $C O_{2}$ $CO_2$ measured at STP?