Question

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

Answer 1

`m_(CaCO_3)=14.1gCaCO_3`

Step-by-step explanation:

Hello,

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

`CaCO_3(s)\rightarrow CaO(s)+CO_2(g)`

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

`n_(CO_2)=(PV_(CO_2))/(RT)=(1atm*3.45L)/(0.082(atm*L)/(mol*K)*298K) \\\\n_(CO_2)=0.141molCO_2`

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

`m_(CaCO_3)=0.141molCO_2*(1molCaCO_3)/(1molCO_2) *(100gCaCO_3)/(1molCaCO_3) \\\\m_(CaCO_3)=14.1gCaCO_3`

Best regards.

Answer 2

15.4 g

Step-by-step explanation:

Let's consider the decomposition of calcium carbonate to form carbon dioxide.

CaCO₃ → CaO + CO₂

Step 1: Calculate the moles 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).

`3.45L * (1mol)/(22.4L) =0.154 mol`

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,

`0.154mol * (100.08g)/(mol) = 15.4 g`