Question

Initially, a test-tube containing a quantity of calcium oxide had a mass of `10.860*g`. This was exposed to moist air to give a final mass of `11.149*g`. What was the starting mass of calcium oxide. and what was the mass of the test-tube?

Answer 1

We need to write a stoichiometric equation to represent the reaction...

Explanation:

`CaO(s) + H_2O(l) rarr Ca(OH)_2(s)`...

Now the initial mass, `10.860*g="mass of CaO+mass of test tube"`

And the final mass,

`11.149*g="mass of CaO+mass of test tube+mass of water"`

And thus `11.149*g-10.860*g=0.289*g="mass of water"`

...and this is a molar quantity of `(0.289*g)/(18.01*g*mol^-1)=0.0161*mol`...and thus there were....

`0.0161*molxx56.08*g*mol^-1=0.900*g` with respect to `CaO`.

And therefore the mass of the test tube is `{10.860-0.900}*g=9.96*g`