How many grams of calcium carbonate will I need to form 3.45 liters of carbon dioxide.

Since the conditions are not explicit, we will suppose that CO₂ is at standard temperature and pressure (STP). In these conditions, 1 mole of any gas has a volume of 22.4 L (assuming ideal behavior).

[tex]3.45L \times \frac{1mol}{22.4L} =0.154 mol[/tex]

Step 2: Calculate the moles of calcium carbonate

The molar ratio of CaCO₃ to CO₂ is 1:1. Then, the moles of CaCO₃ required are 0.154 moles.

Step 3: Calculate the mass of calcium carbonate

The molar mass of CaCO₃ is 100.09 g/mol. Then,

[tex]0.154mol \times \frac{100.08g}{mol} = 15.4 g[/tex]

sebassandin sebassandin · Answer 2 · 2020-05-19T02:35:17+02:00

Answer:

[tex]m_{CaCO_3}=14.1gCaCO_3[/tex]

Explanation:

Hello,

In this case, we notice the following chemical reaction must be considered:

[tex]CaCO_3(s)\rightarrow CaO(s)+CO_2(g)[/tex]

Thus, for the carbon dioxide at 1 atm and 298K, the produced moles, with the given formed volume, result:

[tex]n_{CO_2}=\frac{PV_{CO_2}}{RT}=\frac{1atm*3.45L}{0.082\frac{atm*L}{mol*K}*298K} \\\\n_{CO_2}=0.141molCO_2[/tex]

Next, we use the stoichiometry and the molar mass of calcium carbonate (100g/mol) to compute its required grams:

[tex]m_{CaCO_3}=0.141molCO_2*\frac{1molCaCO_3}{1molCO_2} *\frac{100gCaCO_3}{1molCaCO_3} \\\\m_{CaCO_3}=14.1gCaCO_3[/tex]

Best regards.