The given chemical reaction given above is already balanced such that the number of atoms in the left hand side of the equation is equal to that of the right hand side. Using the dimensional analysis, proper conversion factors and the molar masses,
mass of nitrogen = (0.129 g H₂)(1 mol H₂/2 g H₂)(1 mol N₂/3 mol H₂)(28 g N₂/1 mol N₂)
mass of nitrogen = 0.602 g N₂
Therefore, 0.602 g of nitrogen will be required for he reaction.
Answer:
0.21mol Ar (g)
Explanation:
To convert from litres to moles at STP we must divide the amount of litres by 22.4.
4.7 / 22.4 = 0.21mol Ar (g)
N₂H₄ :
N = - 2
H = + 1
2 * ( -2 ) + ( 1 * 4 ) = - 4 + 4 = 0
hope this helps!
Yes it is C and B because I did this before.
