Question

Not a timed or graded assessment. Please show calculation work. We already know the limiting reactant is CaCO3 and the maximum mass of calcium chloride is 11.1g. Question prompt “what is the maximum volume of carbon dioxide CO2 that can form if the reaction is performed at STP”. Quick answer = amazing review :)

Answer 1

22.4 L of carbon dioxide CO2.

Step-by-step explanation:

We need to know what is the maximum volume of carbon dioxide, so here are the steps:

The chemical equation is unbalanced because in the reactants we have 1 mol of hydrogen and in the products, we have 2 moles of hydrogen. The balanced equation would be:

`CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2.`

Now, as we know that the limiting reactant is CaCO3, let's find the number of moles of CaCO3. We have to convert 10.0 g CaCO3 to moles using its shown molar mass (100.086 g/mol):

`10.0\text{ g CaCO}_3\cdot\frac{1\text{ mol CaCO}_3}{100.086\text{ g CaCO}_3}=0.999\text{ moles CaCO}_3\approx1.00\text{ mol CaCO}_3.`

And with this number, we can find the number of moles of CO2 produced. In the chemical equation, you can see that we have 1 mol of CaCO3 reacted and produces 1 mol of CO2, so the molar ratio between CaCO3 and CO2 is 1:1. This means that 1.00 mol of CaCO3 obtained, will produce 1.00 mol of CO2.

The final step is to convert from moles to liters. To find this, we're going to use the STP conditions. In this case, the standard volume of STP is 22.4 L/mol (There are 22.4 L in 1 mol). The conversion is:

`1.00\text{ mol CO}_2\frac{22.4\text{ L CO}_2}{1\text{ mol CO}_2}=22.4\text{ L CO}_2.`

We're going to have 22.4 L of carbon dioxide CO2.