Question

Crystals of washing soda, Na₂CO₃.10H₂O, lose most of their water content when left in an open container.

Na₂CO₃.10H₂O(s)  → Na₂CO₃.H₂O(s)  +  9 H₂O(g)

What will be the decrease in mass for 100.0 g of the crystals from this process?

a. 43.4g

b.56.6 g

c. 63.0 g

d. 81.8 g

Answer 1

Molar mass of Na2CO3 = 106 g/mol

Molar mass of H2O = 18 g/mol

So

Na2CO3.10 H2O = 106 + 180 = 286 g/mol

Na2CO3.H2O = 106 + 18 = 124 g/mol

100 g of Na2CO3.10 H2O = 100 g/286 g/mol => 0.3496 moles

1 mol of Na2CO3.10H2O produces 1 mol of Na2CO3.H2O and 9 moles of H2O

Therefore 0.3496 moles of Na2CO3.10H2O could produce 9*0.3496 moles of H2O ( 3.146 moles)

mass of 3.146 moles H2O = 3.146 mol * 18 g/mol => 56.64 g of H2O.

mass of 1*0.3496 moles Na2CO3.H2O = 0.3496 mole*124 g/mol => 43.4 g of Na2CO3

After loosing H2O ,the resulting mass of Na2CO3.H2O is 43.4 g

So Answer is option A) 43.4 g