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
70
1 mmol sodium = 23 mg sodium chloride 80 mmol sodium chloride = 80 x 23= 1840 mg sodium chloride
234 grams
The number of atoms in 7.5 grams of sodium chloride is 1.55X10^23. Sodium chloride is more commonly known as salt or table salt.
257g
75 g sodium chloride contain 29,75 g sodium.
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.
70
266,86 g aluminium chloride are obtained.
58.9g
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
2.472686481
From the law of conservation of mass: 58.5 - 23 = 35.5 grams of chlorine required. Interestingly, 23 and 35.5 are approximations to the gram atomic masses of sodium and chlorine respectively.
Sodium chloride got its name because it's chemical formula is NaCl; 1 sodium atom and one chlorine atom. When a chlorine atom is part of an ionic compound, it's end is changed from -ine to -ide.
The atomic or ionic mass of sodium is 22.99, and the atomic or ionic mass of chlorine is 35.45. A formula unit of sodium chloride contain one ion of each. Therefore, the mass ratio between sodium and chlorine in sodium chloride is 0.649. The mass ratio between 46 and 70 is 0.657. Therefore, chloride is the limiting reactant in this pair. The mass ratio of chloride to sodium chloride is 35.54/(22.99 + 35.54) or 0.607. Therefore, the mass of sodium chloride formed will be 70/(0.607) or 115 gm, where the depressed last digit indicates that it may not be accurate to + 1. (The limiting datum, 70, has only two significant digits.)
23.3772 grams are there in four tenths moles of sodium chloride
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