Question #3f1d9

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Question #3f1d9

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Answer 1 · 2014-04-04T02:09:38

For this solution, since the reaction is not at Standard Temperature and Pressure (STP), you will need to use the Ideal Gas Law to solve for moles of $C O_{2}$ $CO_2$ and then Stoichiometry to solve for the grams of $C a C O_{3}$ $CaCO_3$ .

To begin with we find the knowns and unknowns for the Ideal Gas Law, PV = nRT.

P = 1.00 atm
V = 3.45 L
n = ??
R = 0.0821 $\frac{a t m L}{m o l K}$ $(atm L)/(mol K)$
T = 298 K

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

$n = \frac{(1.00 a t m) (3.45 L)}{(.0821 \frac{a t m L}{m o l K}) (298 K)}$ $n = ((1.00 atm)(3.45 L))/((.0821 (atmL)/(molK))(298K))$

n = 0.14 moles $C O_{2}$ $CO_2$

Now you can use the balanced chemical equation and stoichiometry to solve for the grams of $C a C O_{3}$ $CaCO_3$

moles $\to$ $->$ moles $\to$ $->$ grams

$0.14 m o l C O_{2} x \frac{1 m o l C a C O_{3}}{1 m o l C O_{2}} x \frac{100.0869 g C a C O_{3}}{1 m o l C a C O_{3}} = 14.01 g C a C O_{3}$ $0.14 mol CO_2 x (1 mol CaCO_3)/(1 mol CO_2) x (100.0869 g CaCO_3)/(1 mol CaCO_3) = 14.01 g CaCO_3$

I hope this was helpful.
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Question #3f1d9

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