The needed mass of sodium is 91,659 g (for 2 moles H2).
The molar mass of sodium hydrogen carbonate is 84 grams per mole, therefore 0.5 moles of it weighs 42 grams.
It will be 11.5 gm od Na metal
Balanced equation first. 2Na + Cl2 -> 2NaCl 35 grams NaCl (1 mole NaCl/58.44 grams )( 1 mole Cl2/2 mole NaCl )( 70.9 grams Cl2/1 mole Cl2) = 21 grams of Cl2 needed
0,257 g hydrogen are needed.
91.88g
Sodium sulfate is not prepared from hydrogen chloride.
12.01 g NH3 = 0.667 mol NH3 = 1 mol H2 = 2.0 g H2
The molar mass of sodium hydrogen carbonate is 84 grams per mole, therefore 0.5 moles of it weighs 42 grams.
It will be 11.5 gm od Na metal
None. There is no hydrogen in sodium
200 g sodium acetate contain 8,92 g hydrogen.
i think that you have to add 100 gramd to produce the amount of 900 grams
Balanced equation first. 2Na + Cl2 -> 2NaCl 35 grams NaCl (1 mole NaCl/58.44 grams )( 1 mole Cl2/2 mole NaCl )( 70.9 grams Cl2/1 mole Cl2) = 21 grams of Cl2 needed
70
molecular formula for sodium chloride = NaClIf the mole (n) for NaCl = 5.3 moles, then the mole of sodium (Na) = 5.3 moles as well. 1 to 1 ratio mass = moles X molar mass m = 5.3 x 22.9 = 121.37 grams of sodium in 5.3 moles of sodium chloride
The balanced equation for the formation of NH3 is N2 + 3 H2 --> 2 NH3. 13.64 grams of ammonia is equal to .801 moles. Then 1.2015 moles of hydrogen are needed, or 2.42 grams.
155.2 g