sjccengagenowcom OWLv2 Online teaching and learning resourc

sjc.cengagenow.com OWLv2 | Online teaching and learning resource from Cengage Learning cakcium carbonate (s) calcium oxide (s) + carbon dioxide (g) balance [Review Topics] Use the References to access important values if needed for this question. How many moles of calcium carbonate are needed to produce 14.9 L of carbon dioxide according to the following reaction at 0 °C and 1 atm? calcium carbonate (s) calcium oxide (s) + carbon dioxide(g) moles calcium carbonate 2req Submit Answer Retry Entire Group 9 more group attempts remaining 2req s 2req s 2req ts 2req ts 2req Next

Solution

Answer:-

Therefore the balance equation is,

CaCO3(s) CaO(s) + CO2(g)

Where indicates high heat, and CaCO3 (calcium carbonate) has decomposed into CaO(s) (calcium oxide) and CO2(g) (gaseous carbon dioxide).

You can start by assuming CO2(g) is an ideal gas. At STP, we are at a temperature of 0C, or 273.15 K, and 1 atm of pressure. We know the volume produced is 14.9 L of gas.

So we have to determine the volume in 1 mol (the molar volume) of ideal gas to figure out how many moles of CO2 we made.

Therefore the equation of ideal gas is,

PV= nRT

Where,

P = the pressure (atm),

V = the volume (L),

n = the number of mols of ideal gas,

R = the universal gas constant (we choose 0.082057 Latm/molK because we are using 1 atm for pressure and L for volume),

T = the temperature in K,

Respectively,rearrange to get the molar volume, V/n:

V / n=RT / P

= [(0.082057 Latm/molK)(273.15 K)] / 1 atm

= 22.414 L

So, the moles of CO2 we have is:

14.9L × (1 mol / 22.414L) = 0.6648 moles CO2

So, back-calculations will give the moles of CaCO3 needed,

0.6648 moles CO2× [ (1 mol CaCO3(s)) / (1 mol CO2(g))]

= 0.6648 moles CaCO3(g)