Question

The example given the following problem:

A step-by-step example of this process, using the balanced equation from Figure 1, is shown below: CuSO4(aq)2NaOH(aq) -> Cu(OH)2(s) Na2SO4(aq) Assuming there are only 5.70 grams of CuSO4 available, how many grams of NaOH are necessary to rea stoichiometric quantities? How many grams of solid Cu(OH)2 are expected to be produced? Step 1) Check to ensure that the equation is balanced. To do this, ensure that there is the same number o atoms from each element on both sides of the equation Step 2) Convert the 5.70 grams of CuSO4 to moles of CuSO4 1 mol CuSO4 159.62 g CuSO4 0.0357 mol CuSO4 5.70 g CuSO4 x Step 3) Evaluate the molar ratio between CuSO4 and NaOH 2 mol NaOH 0.0357 mol CUSO4 x 0.0714 mol NaOH 1 mol CuSO4 The chemical equation states that for 1 mole of CuSO4, 2 moles of N2OH are needed for stoichiometric quantities. Using the information calculated in step 2, if there are 0.0357 moles of CuSO4, then 0.0714 moles of NaOH are required for a complete reaction.

I used 1.50 grams of the CaCl₂•2H₂O.

I just want to make sure I did my calculations right? Could someone please review my answers and provide appropriate feedback my their answers to mine differ?

If they do, can you show your work.

Answer 1

1- The reaction between CaCl2.2H2O and Na2CO3 is-

Na₂Co₃(aq) + CaCl₂*2H₂O(aq) --------------> CaCo₃(s) + 2 NaCl(aq) + 2 H₂O

Or we can also write this as-

Na₂Co₃(aq) + CaCl₂(aq) --------------> CaCo₃(s) + 2 NaCl(aq)

It indicates that to get 1 mole of CaCo₃(s) we need to react 1 mole Na₂Co₃(aq) and 1 mole of CaCl₂(aq)

a-

Now given amount of CaCl₂*2H₂O(aq) taken = 1.50 g

The molar mass of CaCl₂*2H₂O(aq) is = 147.0146 g/mol

Thus mols of CaCl₂*2H₂O(aq) taken = mass / molar mass

= 1.50 g / 147.0146 g/mol

= 0.01 mols

Similarly If we consider only CaCl2 moles, then in 1 mole of CaCl₂*2H₂O(aq), we have 1 mole of CaCl2. Then in 0.01 mols of CaCl₂*2H₂O(aq), we have 0.01 mole of CaCl2.

b-

Now

From the 1st reaction, we can see for complete reaction of 1 mole of CaCl₂*2H₂O(aq), we need 1 mole of Na₂Co₃(aq)

Thus for complete reaction of 0.01 mole of CaCl₂*2H₂O(aq), we will need 0.01 mole of Na₂Co₃(aq)

c-

Thus mass of 0.01 mole of Na₂Co₃(aq) = number of moles of Na₂Co₃(aq) * molar mass of Na₂Co₃(aq)

= 0.01 mole * 105.9 g/mol

= 1.059‬ g

Thus after final reaction of 0.01 mole of CaCl₂*2H₂O with 0.01 mole of Na₂Co₃, we wil get 0.01 moles CaCo₃(s)

d-

Again

mass of 0.01 mole of CaCo₃(s) = number of moles of CaCo₃(s) * molar mass of CaCo₃(s)

= 0.01 mole * 100 g/mol

= 1 g

e-

Now the experimental yeild = 0.7 g

Thus % yeild = experimental yeild/ theoritical yeild * 100

= 0.7g/ 1 g * 100

= 70 %