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?

1 Answer
Sep 1, 2016

Answer:

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#