2.472686481
To find the answer, we multiply the 7 grams of NaCl by the ratio of the molar mass of chlorine over the molar mass of sodium chloride. By doing this, we find that there are about 4.25 grams of chlorine in 7 grams of NaCl.
142 grams Cl2
The formula of aluminium chloride is AlCl3. The atomic weight of aluminium is 27 and that of chlorine is 35.5. That means 35.5*3 grams of chlorine will combine with 27 grams of aluminium. So 33 grams of chlorine will combine with 8.37 grams of aluminium. The addition of both makes it 41.37 grams. In this reaction, the whole chlorine will be utilized and only part of the aluminium.
You have one mole sodium and one mole chlorine so one mole NaCl will be formed that weight as (23+35.5)=58.5 g
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
75 g sodium chloride contain 29,75 g sodium.
6
To find the answer, we multiply the 7 grams of NaCl by the ratio of the molar mass of chlorine over the molar mass of sodium chloride. By doing this, we find that there are about 4.25 grams of chlorine in 7 grams of NaCl.
Balanced equation. 2Na + Cl2 >> 2NaCl 46 grams sodium = 2 mol 23 grams Chlorine = 0.65 mol ( I think Chlorine is limiting ) 0.65 mol Cl (2mol Na/1mol Cl ) = 1.3 mol ( you do not have that; Cl limits ) 0.65 mol Cl (2mol NaCl/1mol Cl2 )(58.44g/1mol NaCl ) = 75.9 grams
142 grams Cl2
The formula of aluminium chloride is AlCl3. The atomic weight of aluminium is 27 and that of chlorine is 35.5. That means 35.5*3 grams of chlorine will combine with 27 grams of aluminium. So 33 grams of chlorine will combine with 8.37 grams of aluminium. The addition of both makes it 41.37 grams. In this reaction, the whole chlorine will be utilized and only part of the aluminium.
75 grams per mole
70
You have one mole sodium and one mole chlorine so one mole NaCl will be formed that weight as (23+35.5)=58.5 g
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
384.5g
5.0 grams.