Question

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

Answer 1

Answer:

$m_{CaCO_3}=14.1gCaCO_3$

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:

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

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

$m_{CaCO_3}=0.141molCO_2*\frac{1molCaCO_3}{1molCO_2} *\frac{100gCaCO_3}{1molCaCO_3} \\\\m_{CaCO_3}=14.1gCaCO_3$

Best regards.

Answer 2

Answer:

15.4 g

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).

$3.45L \times \frac{1mol}{22.4L} =0.154 mol$

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,

$0.154mol \times \frac{100.08g}{mol} = 15.4 g$