Suppose you start with 1.80 g of mgco3·5h2o, after complete dehydration how many grams of the sample do you have remaining?

1 answer:

5 0

The number of grams of the sample that remained after hydration is 0.87 grams

calculation
calculate the molar mass of MgCO3. 5 H2O which is

24 + 12+ (3 x16) + ( 5 x18) = 174 g/mol

we well know after hydration water evaporate leaving behind MgCO3
therefore the mass remained is for MgCO3

find the molar mass of MgCO3 = 24 +12 +( 3 x16) = 84 g/mol

the mass remained is therefore

= molar mass of MgCO3 / molar mass of MgCO3.5H2O x 1.80 g

= 84g/mol / 174 g/mol x1.8g = 0.87 grams

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