Question

A sample of calcium nitrate, `Ca(NO_3)_2`, with a formula weight of 164 g/mol, has `5.00 x 10^25` atoms of oxygen. How many kilograms of `Ca(NO_3)_2` are present?

Answer 1

There are 2.27 kg of `"Ca"("NO"_3)_2` present.

Explanation:

Step 1. Calculate the moles of `"O"` atoms.

`"Moles of O" = 5.00 × 10^25 color(red)(cancel(color(black)("atoms O"))) × "1 mol O"/(6.022 × 10^23color(red)(cancel(color(black)("atoms O")))) = "83.03 mol O"`

Step 2. Calculate the moles of `"Ca"("NO"_3)_2`.

The formula tells us that 1 mol of `"Ca"("NO"_3)_2` contains 6 mol of `"O"` atoms.

∴ `"Moles of Ca"("NO"_3)_2 = 83.03color(red)(cancel(color(black)( "mol O"))) × ("1 mol Ca"("NO"_3)_2)/(6 color(red)(cancel(color(black)("mol O")))) = "13.84 mol Ca"("NO"_3)_2`

Step 3. Calculate the mass of `"Ca"("NO"_3)_2`

`"Mass of Ca"("NO"_3)_2 = 13.84 color(red)(cancel(color(black)("mol Ca"("NO"_3)_2))) × ("164 g Ca"("NO"_3)_2)/(1 color(red)(cancel(color(black)("mol Ca"("NO"_3)_2)))) = "2270 g Ca"("NO"_3)_2 = "2.27 kg Ca"("NO"_3)_2`