Na2CO3 + CaCl2•2H2O -> CaCO3 + 2NaCl + 2H2O Calculate how many moles of CaCl2•2H2O are present in 1.50 g of CaCl2•2H2O and then calculate how many moles of pure CaCl2 are present in 1.50 g of CaCl2•2H2O. Then determine how many grams of Na2CO3 are necessary to reach stoichiometric quantities.
For CaCl2 I got 0.0135 mol but I have seen some put 0.0102 mol. Which is it?
For the initial mol of Na2CO3 I got 0.0102 mol but again I’m not sure if I’m right.
For the grams of Na2CO3 I got 1.08 g
Can someone help me figure out if I have this correct?

See explanation

Explanation:

Number of moles = reacting mass/molar mass

Number of moles of CaCl2•2H2O = 1.50 g/147.02 = 0.0102 moles

From the equation;

Na2CO3 + CaCl2•2H2O -> CaCO3 + 2NaCl + 2H2O

We can see is 1:1

1 mole of Na2CO3 reacts with 1 mole of CaCl2•2H2O

x moles of Na2CO3 reacts with 0.0102 moles of CaCl2•2H2O

x = 1 × 0.0102 moles/1

x = 0.0102 moles of Na2CO3

Mass of Na2CO3 = 0.0102 moles of Na2CO3 × 106g/mol = 1.08 g of Na2CO3

i just calculated it all out, and not including weekends, and all the different days off. as of this school year (2018-2019) students are spending 2198 hours in school this year

the answer would be a