Question

Given the balanced reaction above, if 0.350 g of CaCO3 is reacted with 45 mL of 1.5 M HCl then which reagent is the limiting reagent and what mass of carbon dioxide should be produced in theory?Complete and balance the following exchange (Hint one of the products is CO2)CaCO3(s) + HCl (aq) →

Answer 1

`\begin{gathered} CaCO_3\text{ is the limiting reagent} \\ 0.154\text{ g of CO}_2\text{ would be produced in theory} \end{gathered}`

Step-by-step explanation:

We start by completing the balanced equation of the reaction.

We have that as:

`CaCO_3\text{ + 2HCl }\rightarrow\text{ CaCl}_2\text{ + H}_2O\text{ + CO}_2`

Now, we want to get the limiting reagent

The limiting reagent is the reagent that produces less amount of the product

Now, let us get the number of moles of CaCO3 that reacted

That would be the mass of CaCO3 given, divided by the molar mass of CaCO3

The molar mass of CaCO3 is 100 g/mol

Thus, the number of moles of CaCO3 that reacted will be:

`(0.350)/(100)\text{ = 0.0035 mol}`

Now, let us get the number of moles of CO2 produced

That would be 0.0035 mol too since the mole ratio of CaCO3 to CO2 in the balanced equation of reaction is 1:1

We proceed to get the number of moles of HCl that reacted

We can get this by multiplying the given molarity by the volume in liters

We have that as:

`1.5\text{ }*(45)/(1000)\text{ = 0.0675 mol}`

From here, the number of moles of CO2 produced is half, since the mole ratio is 1 to 2

The number of moles of CO2 produced will be:

`(0.0675)/(2)\text{ = 0.03375 mol}`

The number of moles of CO2 produced by CaCO3 is lesser, and that means it is the limiting reagent

Now, to get the mass of CO2 produced in theory, we multiply the number of moles by the molar mass of CO2

The molar mass of CO2 is 44 g/mol

Thus, we have the mass that was produced as:

`44\text{ }*\text{ 0.0035 = 0.154 g}`