A 8.85 g sample of hydrate of sodium carbonate, Na2CO3 ∙ XH2O was heated carefully until no more mass was lost from the sample. After heating, the final weight of the material was 7.55 g. How many moles of water left the compound? 0 moles

The compound is Na2CO3*1H2O

There left 0.0721 moles H2O

Explanation:

Step 1: Data given

Mass of Na2CO3* XH2O = 8.85 grams

After heating the mass = 7.55 grams

Molar mass of H2O = 18.02 g/mol

Molar mass of Na2CO3 = 105.99 g/mol

Step 2: Calculate mass of water lost

8.85 - 7.55 = 1.30 grams

Step 3: Calculate moles Na2CO3

Moles Na2CO3 = mass Na2CO3 / molar mass Na2CO3

Moles Na2CO3 = 7.55 grams / 105.99 g/mol

Moles Na2CO3 = 0.0712 moles

Step 4: Calculate moles H2O

Moles H2O = 1.30 grams / 18.02 g/mol

Moles H2O = 0.0721

Step 5: Calculate mol ratio

We divide by the smallest amount of moles

Na2CO3: 0.0712 / 0.0712 = 1

H2O : 0.0721 / 0.0712 = 1

This means for 1 mol Na2CO3 we have 1 mol H2O

The compound is Na2CO3*1H2O

There left 0.0721 moles H2O

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2