If the reaction below proceeds with a 96.8% yield, how many kilograms of CaSO4 are formed when 5.97 kg SO2 reacts with an excess of CaCO3 and O2?
2 CaCO3(s) + 2 SO2(g) + O2(g) 2 CaSO4(g) + 2 CO2(g)
first convert the kg of SO2 to grams of SO2 then convert to moles by dividing by its molar mass
secondly convert between moles of 2SO2 and 2CaSO4 which is a 2/2 ration so no change in moles
then multiply your moles by CaSO4's molar mass to get grams, then multiply that amount by 96.8% (.968) to get a 96.8% yeild of CaSO4's grams and convert to kg by dividing by 1000
Well, that's a very "weigh"-ty question! Let's calculate it step by step.
First, we need to determine the moles of SO2 you have. Since we know the molar mass of SO2 is 64.06 g/mol, we can divide the mass (5.97 kg) by the molar mass to find the moles of SO2.
5.97 kg SO2 × (1000 g / 1 kg) / 64.06 g/mol = 93.22 mol SO2
Next, we'll use the balanced equation to find the moles of CaSO4 formed. According to the equation:
2 moles of SO2 produce 2 moles of CaSO4
So, if we have 93.22 moles of SO2, we'll end up with the same amount (93.22 moles) of CaSO4.
Finally, we'll convert the moles of CaSO4 into kilograms using its molar mass (136.14 g/mol).
93.22 mol CaSO4 × 136.14 g/mol × (1 kg / 1000 g) = 12.7 kg CaSO4
So, approximately 12.7 kilograms of CaSO4 will be formed in this reaction with a 96.8% yield. Calcium So-fourtunately, I can't make the yield any higher!
To find the mass of CaSO4 formed, we need to determine the limiting reactant first, then use stoichiometry to calculate the mass of the product.
1. Calculate the molar masses:
- CaCO3: 40.08 g/mol + 12.01 g/mol + (3 * 16.00 g/mol) = 100.09 g/mol
- SO2: 32.07 g/mol + (2 * 16.00 g/mol) = 64.07 g/mol
- O2: (2 * 16.00 g/mol) = 32.00 g/mol
- CaSO4: 40.08 g/mol + 32.07 g/mol + (4 * 16.00 g/mol) = 136.15 g/mol
2. Convert the mass of SO2 to moles:
5.97 kg * (1000 g/kg) / 64.07 g/mol = 93.24 mol SO2
3. Determine the limiting reactant:
Divide the number of moles of each reactant by their stoichiometric coefficients to find the mole ratio:
- CaCO3: 0.5 * (93.24 mol SO2) = 46.62 mol CaCO3
- O2: 0.5 * (93.24 mol SO2) / 2 = 23.31 mol O2
The number of moles of CaCO3 is less than that of O2, so CaCO3 is the limiting reactant.
4. Calculate the mass of CaSO4 formed using the moles of CaCO3:
46.62 mol CaCO3 * (1 mol CaSO4 / 2 mol CaCO3) * (136.15 g/mol) = 3172.69 g = 3.17269 kg
5. Calculate the actual yield, taking into account the 96.8% yield:
3.17269 kg * 0.968 = 3.07412 kg
Therefore, approximately 3.07 kg of CaSO4 are formed when 5.97 kg SO2 reacts with an excess of CaCO3 and O2.
To determine the amount of CaSO4 formed, we need to follow these steps:
1. Balance the equation: The given equation is already balanced.
2. Convert the mass of SO2 to moles: The molar mass of SO2 is 64.06 g/mol. So, divide the given mass by the molar mass to obtain the number of moles:
Moles of SO2 = mass of SO2 / molar mass of SO2
Moles of SO2 = 5.97 kg / 64.06 g/mol
3. Determine the limiting reagent: To find the limiting reagent, compare the number of moles of each reactant with the stoichiometric coefficients. In this case, SO2 is the limiting reagent because it has a stoichiometric coefficient of 2, while CaCO3 and O2 each have a stoichiometric coefficient of 1.
4. Calculate the moles of CaSO4 formed: Since the reaction has a 96.8% yield, multiply the moles of SO2 by the stoichiometric coefficient of CaSO4:
Moles of CaSO4 = Moles of SO2 * (1 mol CaSO4 / 2 mol SO2)
5. Convert moles of CaSO4 to mass: Multiply the moles of CaSO4 by its molar mass to obtain the mass of CaSO4 formed:
Mass of CaSO4 = Moles of CaSO4 * molar mass of CaSO4
6. Convert the mass of CaSO4 from grams to kilograms, if required.
Now, let's substitute the values and calculate the mass of CaSO4 formed:
Moles of SO2 = 5.97 kg / 64.06 g/mol
Moles of CaSO4 = Moles of SO2 * (1 mol CaSO4 / 2 mol SO2)
Mass of CaSO4 = Moles of CaSO4 * molar mass of CaSO4
It's important to note that you should use the accurate molar masses for each element or compound involved in the calculation.