Answer to Question #114238 in General Chemistry for Daniel

Question

If 1.000 x 10^23 atoms of copper reacted with nitric acid, how many grams of water would be produced?

Answer 1

$4HNO_3(l)+Cu(s) \rightarrow Cu(NO_3)_2(aq) + 2NO_2(g)+2H_2O(l)$

Convert number of atoms of copper to moles

$1.000\times 10^{23} (\frac{1 mole}{6.022\times10^{23}})=0.16606 moles$

Convert moles of copper to moles of water:

$0.16606 moles Cu (\frac{2 moles H_2O}{1 moles Cu})=0.33212 moles H_2O$

Convert moles of water to mass of water

$0.33212 moles H_2O (\frac{18.0.15 g H_2O}{1 mole H_2O})= 5.983 g H_2O$