75 g sodium chloride contain 29,75 g sodium.
If 75 grams of sodium reacts with 25 grams of chlorine how many grams of salt will be produced?
2.472686481
The answer is o,5 mol.
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
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
Sodium has an atomic weight of 22.99 g/mol. Chlorine has an atomic weight of 35.45 g/mol. NaCl has an atomic weight of 58.44 g/mol. Therefore 92g of sodium would yield 233.86g of sodium chloride (NaCl).
384.5g
6
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
The balanced chemical equation for the reaction between calcium and chlorine gas to produce calcium chloride is: Ca + Cl2 -> CaCl2. From this equation, we can see that one mole of calcium reacts with one mole of chlorine gas to produce one mole of calcium chloride. The molar mass of calcium is 40.08 g/mol and the molar mass of chlorine gas is 70.90 g/mol. This means that 10.0 grams of calcium is equivalent to 0.249 moles of calcium and 20.0 grams of chlorine gas is equivalent to 0.282 moles of chlorine gas. Since the ratio of calcium to chlorine gas in the balanced chemical equation is 1:1, this means that 0.249 moles of calcium would react completely with 0.249 moles of chlorine gas, leaving an excess of 0.033 moles (or 2.34 grams) of chlorine gas. The limiting reactant in this reaction is calcium, and the maximum amount of calcium chloride that can be produced is equivalent to the number of moles of the limiting reactant, which is 0.249 moles (or 27.8 grams) of calcium chloride.