Given the reaction 2HI (aq) + CaCO3(s) = CaI2(aq)+CO2(g)+H2O(l) What volume of of CO2 can be produced from 255mL of 3.0M HI and 75.2g of CaCO3 at STP?

75.2g CaCO3/100.09g/mol CaCO3 =.751mol CaCO3

(3M)(.255L)=.765 mol HI

This is a limiting reagent (LR) problem. You know that because amounts are given for BOTH reactants.

Your first step is ok.

Using the coefficients in the balanced equation convert mols CaCO3 to mols CO2.
Do the same and convert mols HI to mols CO2.
It is likely that these two values will not be the same; the correct value in LR problems is ALWAYS the smaller value and the reagent responsible for that number is called the LR.

Now convert the the LR value in mols to grams CO2. g = mols x molar mass = ?

Okay so this was my next step

0.765 mol CaCO3 x 1 mol CO2/2 mol HI = 0.3825mol CO2.

At STP, 1 mol=22.4L so (22.4)(0.3825mol)= 8.568 L CO2

Does that look right?

First, we need to determine the limiting reactant in the reaction. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed.

Using the stoichiometry of the balanced equation, we can determine the moles of CO2 produced from the moles of HI and CaCO3.

From the balanced equation, we can see that it takes 2 moles of HI to produce 1 mole of CO2.
Therefore, from 0.765 mol HI, we can calculate the moles of CO2 produced:

0.765 mol HI × (1 mol CO2 / 2 mol HI) = 0.3825 mol CO2

Next, we can use the ideal gas law to calculate the volume of CO2 produced at STP (Standard Temperature and Pressure).

STP conditions are defined as a temperature of 273.15 K (0°C) and a pressure of 1 atmosphere (1 atm).
The molar volume of gas at STP is approximately 22.4 L/mol.

Using the molar volume of gas, we can calculate the volume of CO2 produced:

0.3825 mol CO2 × 22.4 L/mol = 8.5675 L CO2

Therefore, the volume of CO2 that can be produced from 255 mL of 3.0 M HI and 75.2 g of CaCO3 at STP is 8.5675 L.

To determine the volume of CO2 produced, we need to use stoichiometry and the ideal gas law.

First, let's examine the balanced equation:
2HI(aq) + CaCO3(s) → CaI2(aq) + CO2(g) + H2O(l)

From the equation, we can see that 1 mole of CaCO3 produces 1 mole of CO2. So, we need to find the number of moles present in 75.2 grams of CaCO3.
Using the molar mass of CaCO3:
75.2 g CaCO3 / 100.09 g/mol CaCO3 = 0.751 mol CaCO3

Now, let's consider the reaction between HI and CaCO3. From the equation, we can see that 2 moles of HI react with 1 mole of CaCO3 to produce 1 mole of CO2. So, we need to find the number of moles of HI present in the given volume and concentration.
Using the given concentration and volume:
(3.0 M)(0.255 L) = 0.765 mol HI

Now, we have found both the number of moles of CaCO3 (0.751 mol) and HI (0.765 mol). Since CaCO3 is the limiting reactant (the one that limits the formation of product), we will use its amount to determine the amount of CO2 produced.

Since the reaction occurs at STP (Standard Temperature and Pressure), we can use the ideal gas law equation to find the volume of CO2 produced.

The ideal gas law equation is: PV = nRT

At STP, the pressure (P) is 1 atm, the temperature (T) is 273.15 K, and the ideal gas constant (R) is 0.0821 L·atm/(mol·K).
We can substitute the known values into the equation: (1 atm)(V) = (0.751 mol)(0.0821 L·atm/(mol·K))(273.15 K)

Now, solve for V (volume of CO2):
V = (0.751 mol)(0.0821 L·atm/(mol·K))(273.15 K) / 1 atm

Calculating this equation will give you the volume of CO2 produced in liters.