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# 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?

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

**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##