NaH(s)+ H2O (l)=>NaOH(aq)+H2(g)
You want to calculate the mass of NaH, I assume. Otherwise, the question isn't clear. It simply says calculate the mass(??)
So, calculate the moles of H2 gas that satisfy the conditions of 982 ml at 28ºC and 765 torr. But you must subtract the vapor pressure of water at 28º to get the actual pressure of the H2 gas. So, the actual conditions are 982 ml (0.982 L) and 301 K and 765-28 = 737 torr.
PV = nRT
n = PV/RT = (737 torr)(0.982 L)/(62.4 L-torr/Kmol)(301 K)
n = 0.0385 moles H2
moles NaH needed = 0.0385 moles H2 x 1 mole NaH/mole H2 = 0.0385 moles NaH required
mass of NaH needed = 0.0385 moles x 24 g/mole = 0.925 g NaH
Brainliest Please :)
The scientific notation for 5,098.000 is 5.098000*10^(3).
Here the number also has 7 significant figures.
Hope this helps~
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.
