Assuming you meant cation and not action, gallium would most likely form a cation because it is a group A element
Answer: 16.32 g of 
as excess reagent are left.
Explanation:
To calculate the moles :
According to stoichiometry :
2 moles of 
require = 1 mole of
Thus 0.34 moles of will require=
of
Thus is the limiting reagent as it limits the formation of product and is the excess reagent.
Moles of left = (0.68-0.17) mol = 0.51 mol
Mass of 
Thus 16.32 g of as excess reagent are left.
In order to calculate the mass of nitrogen, we must first calculate the mass percentage of nitrogen in potassium nitrate. This is:
% nitrogen = mass of nitrogen / mass of potassium nitrate
% nitrogen = 14 / 101.1 x 100
The mass of nitrogen = % nitrogen x sample mass
= (14 / 101.1) x 101.1
= 14 grams
The molar weight of nitrogen is 14. Each mole of urea contains two moles of nitrogen. Therefore, for there to be 14 grams of nitrogen, there must be 0.5 moles of urea.
Mass of urea = moles urea x molecular weight urea
Mass of urea = 0.5 x 66.06
Mass of urea = 33.03 grams
0.075 L * 1.0 M = 0.075 mol HCl
H2 is half of HCl by the coefficients divide 0.075 by 2 and get 0.0375 mol H2 gas
