# Moles Of Anhydrous Salt

Moles Of Anhydrous Salt. In our example, 16 g /. Calculate the number of moles of the anhydrous salt left behind.

X/1 = moles of water found above/ moles of anhydrous salt found above. This is 2×39.1 + 12 + 3×16 = 138.2 g/mole. Convert the mass of the anhydrous salt ( m2) by dividing it by the molar mass of the anhydrous salt.

### Calculate The Ratio Of Moles Of H 2 O To Moles Of Anhydrous Kal(So 4) 2.

Correct formula of hydrate (ask your instructor) 17. What is the mole ratio of salt to water in mo(no3)2.5h2o? The moles of salt should be the smaller of the two values.

### In Our Example, 16 G /.

Sugar is a covalent compound where as salt is an ionic compound,so salt while dissolving in the water splits.this is mainly dependent of the 'total. Grams of the dry salt divided by the molar mass = moles The same way you do with most any other known chemical substance:

### Conclusions Provide A Summary Statement Of Your Work In Lab This Week.

Moles of water lost = 0.0312 mol use 6 value and divide by the molar mass of h 2 o. This will allow you to find the mole ratio. Is used for heating the hydrate is 38.4188 % of 4 ):

### Calculate The Number Of Moles Of The Anhydrous Salt Left Behind.

Al = 1 * 26.98 + cl = 3 * 35.5 = 133.48g/mol. Equation 2 number of moles of anhydrous salt after heating, mol = † mass of anhydrous after heating, g 1 ê ë á ˆ ¯ ˜ 1 mol anhydrous salt mass of 1 mol anhydrous salt, g ê ë á á ˆ ¯ ˜ ˜ = 1.060g bacl2 1 ê ë á ˆ ¯ ˜ ¥ 1 mol bacl2 208.3 g bacl2 ê ë It is not too difficult.

### Anhydrous Salt To Be 208.3 G.

First, calculate the number of moles of anhydrous salt are present in 7.93 g. In our example, 0.5 moles of. 16 grams / 160 grams per mole = 0.1 moles the correct percent of the anhydrous ( without water water.