If 0.0490 mol of solid CaCO3 and 380 mL of 0.135 M aqueous H2SO4 are reacted, what volume (L) of gaseous CO2 measured at 1.1 atm pressure and 301 K is produced.

H2SO4(aq) + CaCO3(s) → CO2(g) + CaSO4(s) + H2O(l)

This is a limiting reagent problem. How do I know? Because BOTH reactants are given.

Convert 380 mL of 0.135 M H2SO4 to moles. moles = M x L. You already have moles CaCO3.

Using the coefficients in the balanced equation, convert moles CaCO3 to moles CO2.
Same process, convert moles H2SO4 to moles CO2.
The two answers probably will not agree which means one of them is wrong; the correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
Use that smaller value and PV = nRT to solve for the volume at the conditions listed.

To solve this problem, you can use the concept of stoichiometry. Stoichiometry is a technique that allows us to calculate the quantities of reactants and products involved in a chemical reaction.

First, we need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed and determines the amount of product formed. To find the limiting reactant, we compare the number of moles of each reactant to the stoichiometric ratio in the balanced chemical equation.

The balanced equation is:

H2SO4(aq) + CaCO3(s) → CO2(g) + CaSO4(s) + H2O(l)

From the equation, we see that the stoichiometric ratio between H2SO4 and CO2 is 1:1. Therefore, the number of moles of CO2 produced will be equal to the number of moles of H2SO4.

Given that we have 380 mL of 0.135 M H2SO4, we can calculate the number of moles of H2SO4 using the formula:

moles = concentration (M) × volume (L)

moles of H2SO4 = 0.380 L × 0.135 M = 0.0513 mol

Since we have 0.0490 mol of CaCO3, we can see that H2SO4 is the limiting reactant because we have more CaCO3 than H2SO4. This means that all the H2SO4 will be consumed and converted to CO2.

Now, we need to calculate the volume of CO2 produced. We can do this using the ideal gas law:

PV = nRT

Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)

We are given the pressure (1.1 atm), temperature (301 K), and moles of CO2 (0.0513 mol). Substituting these values into the ideal gas law, we can solve for the volume of CO2:

V = (nRT) / P
V = (0.0513 mol × 0.0821 L·atm/(mol·K) × 301 K) / 1.1 atm
V ≈ 1.463 L

Therefore, the volume of CO2 produced is approximately 1.463 L.