You carefully weigh out 19.00 of powder and add it to 76.95 of solution. You notice bubbles as a reaction takes place. You then weigh the resulting solution and find that it has a mass of 88.16 . The relevant equation is
CaCO3+2HCl ----> H2O + CO2 + CaCl2
Assuming no other reactions take place, what mass of was produced in this reaction?
I have no idea how to attempt to solve this problem, i tried getting moles of carbon but I'm lost. Help please.
Do you not believe in units? Is that 19.00g, 76.95g, and 88.16g? If so, then
19.00g + 76.95g on the left.
Mass H2O and CaCl2 on the right = 88.16g. So the difference must be the mass CO2 on the right.
oops...* Assuming no other rxn's what mass of CO2 was produced in this rxn?
Well, this problem might be a tough nut to crack, but let's try to add a sprinkle of humor to make it more enjoyable!
Alright, let's break it down step by step and see if we can figure it out.
First, we need to find the number of moles of CaCO3. We can do this by dividing the mass of the powder by its molar mass.
But before we do that, let's put on our detective hats and find the molar masses of CaCO3, HCl, H2O, CO2, and CaCl2. Don't worry, we won't need any magnifying glasses for this!
CaCO3: Calcium (Ca) has a molar mass of approximately 40.08 g/mol, Carbon (C) has a molar mass of about 12.01 g/mol, and Oxygen (O) has a molar mass of around 16.00 g/mol. Add them up, and we get a total of 100.09 g/mol.
HCl: Hydrogen (H) has a molar mass of roughly 1.01 g/mol, and Chlorine (Cl) has a molar mass of about 35.45 g/mol. Add them together, and we get a total of 36.46 g/mol.
H2O: Two Hydrogens (H) and one Oxygen (O) make up water (H2O). So, 2(1.01 g/mol) + 16.00 g/mol = 18.02 g/mol.
CO2: Carbon (C) has a molar mass of approximately 12.01 g/mol, and Oxygen (O) has a molar mass of around 16.00 g/mol. Two Oxygen atoms are present in CO2, so 12.01 g/mol + 2(16.00 g/mol) = 44.01 g/mol.
CaCl2: Calcium (Ca) has a molar mass of approximately 40.08 g/mol, and Chlorine (Cl) has a molar mass of about 35.45 g/mol. Multiply the molar mass of chlorine by 2, and we get 2(35.45 g/mol) = 70.90 g/mol. Adding them together, we get 40.08 g/mol + 70.90 g/mol = 110.98 g/mol.
Now that we have our molar masses, let's find the moles of CaCO3 by dividing the mass of the powder (19.00 g) by the molar mass of CaCO3 (100.09 g/mol). This gives us 0.1898 moles of CaCO3.
According to the balanced equation, for every 1 mole of CaCO3, we get 1 mole of CO2. Therefore, we also have 0.1898 moles of CO2.
To find the mass of CO2, we can multiply the moles of CO2 by its molar mass (44.01 g/mol). So, 0.1898 moles × 44.01 g/mol ≈ 8.35 g of CO2.
Therefore, approximately 8.35 grams of CO2 were produced during the reaction.
Voila! We've solved the riddle, and the mystery of the missing reaction product has been unraveled. Well done, Sherlock!
To solve this problem, we need to use stoichiometry and the given information to determine the mass of CO2 produced in the reaction.
Step 1: Calculate the mass of CaCO3:
Given: 19.00 g of powder (CaCO3)
Step 2: Calculate the moles of CaCO3:
To find the moles, we need to use the molar mass of CaCO3, which can be calculated by adding the atomic masses of calcium (Ca), carbon (C), and oxygen (O):
Molar mass of CaCO3 = (40.08 g/mol) + (12.01 g/mol) + (3x16.00 g/mol) ≈ 100.09 g/mol
Moles of CaCO3 = mass of CaCO3 / molar mass of CaCO3 = 19.00 g / 100.09 g/mol
Step 3: Determine the limiting reagent:
The limiting reagent is the one that is completely consumed in the reaction and determines the amount of product formed. In this case, using the balanced equation, we can see that the stoichiometric ratio between CaCO3 and CO2 is 1:1. Therefore, the moles of CaCO3 will be equal to the moles of CO2 produced.
Step 4: Calculate the mass of CO2:
Moles of CO2 = Moles of CaCO3 = 19.00 g / 100.09 g/mol
Mass of CO2 = Moles of CO2 x molar mass of CO2
The molar mass of CO2 can be obtained by adding the atomic masses of carbon (C) and oxygen (O):
Molar mass of CO2 = (12.01 g/mol) + (2x16.00 g/mol) ≈ 44.01 g/mol
Mass of CO2 = (19.00 g / 100.09 g/mol) x 44.01 g/mol
By performing the calculations, you can determine the mass of CO2 produced in the reaction.