For the reactionCaCO3(s)+2HCl(aq)→CaCl2(aq)+CO2(g)+H2O(l)CaCO3𝑠+2HClaq→CaCl2aq+CO2𝑔+𝐻2𝑂𝑙 Calculate the mass of  CO2CO2 that can be obtained if  421421  g of CaCO3CaCO3  and  329329  g of HClHCl  resulted in a 100% yield of CO2CO2 .Molar mass: CaCO3 = 100.09 g mol-1; HCl = 36.46 g mol-1; CO2 = 44.01 g mol-1 199.g199.𝑔 397.g397.𝑔 370.g370.𝑔 467.g467.𝑔 185.g185.𝑔

First, we need to determine the number of moles for both CaCO3 and HCl.

For CaCO3: Number of moles = mass / molar mass = 421 g / 100.09 g/mol = 4.21 mol

For HCl: Number of moles = mass / molar mass = 329 g / 36.46 g/mol = 9.02 mol

From the balanced chemical equation, we can see that the mole ratio of CaCO3 to HCl to CO2 is 1:2:1. This means that for every mole of CaCO3, one mole of CO2 is produced. Similarly, for every 2 moles of HCl, one mole of CO2 is produced.

Therefore, the amount of CO2 that can be produced from 4.21 mol of CaCO3 is 4.21 mol. However, the amount of CO2 that can be produced from 9.02 mol of HCl is 9.02/2 = 4.51 mol.

Since we cannot produce more than the limiting reagent allows, the amount of CO2 produced is limited by the amount of CaCO3, which is 4.21 mol.

Finally, we convert this amount back to grams using the molar mass of CO2:

Mass of CO2 = number of moles * molar mass = 4.21 mol * 44.01 g/mol = 185.2 g

So, the mass of CO2 that can be obtained if 421 g of CaCO3 and 329 g of HCl resulted in a 100% yield of CO2 is approximately 185 g.