19.8-12.6 = g H2O evolv3d = 7.2 g.
12.6/molar mass MnCl2 =
12.6/125.8 = 0.1 mole MnCl2
7.2/18 = 0.4 mol H2O
This is a ratio of 1 MnCl2 to 4 H2O so the formula is
MnCl2*4H2O
A sample of Manganese (II) Chloride has a mass of 19.8 grams before heating, and 12.6 grams after heating until all the water is driven out. Based on this information, how many water molecules were attached to each molecule of Manganese (II) Chloride?
2 answers
3 molecules were attached