Molar Mass NaCl is 58.5g, 39.3% Na (5.90g), Molar Mass Na is 23.0g, 5.90/23.0 = 0.256 moles
Based on the stoichiometry of NaCl, for every one mole of NaCl there is one mole of Na+ and one mole of Cl-. Therefore, there are 2.5 moles Na+ and 2.5 moles Cl-, totaling 5 moles of ions altogether.
Based on the stoichiometry of NaCl, for every one mole of NaCl there is one mole of Na+ and one mole of Cl-. Therefore, there are 1.5 moles Na+ and 1.5 moles Cl-, totaling 3 moles of ions altogether
(7.6 g NaCl) (moles NaCl/58.44 g NaCl) (1 mole Na/1 mole NaCl) ( 22.99 g Na/mole Na) = 2.989 g Na or 3.0 g Na (significant figures) Steps: 1. Change everything to moles: use molar mass/molecular of NaCl 2. Find the molar ratio: since there are 1 Na in every NaCl, then the ratio of NaCl to Na is 1:1 3. Change moles back to grams: use the molar mass/molecular weight of Na 4. Watch out for significant figures! I hope this is right
1.8 moles
(22.99 g/mol Na / 58.43 g/mol NaCl ) x 100 ppm NaCl= 39.35 ppm Na
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Based on the stoichiometry of NaCl, for every one mole of NaCl there is one mole of Na+ and one mole of Cl-. Therefore, there are 2.5 moles Na+ and 2.5 moles Cl-, totaling 5 moles of ions altogether.
Based on the stoichiometry of NaCl, for every one mole of NaCl there is one mole of Na+ and one mole of Cl-. Therefore, there are 1.5 moles Na+ and 1.5 moles Cl-, totaling 3 moles of ions altogether
In each formula unit of NaCl there is one Na+ ion. Therefore, amount of sodium ions = amount of NaCl = 4.25mol
There are 5 formula masses of Na in 5 formula masses of NaCl, as indicated by the fact that the symbol for sodium has no explicit subscript in the formula of NaCl. Neither sodium nor sodium chloride has moles in the strictest sense, since neither of them is covalently bonded.
2 moles of NaCl, of course. Cl would definitely limit in this one to one reaction and you would have 19998 moles Na in excess.
1 mole of NaCl: Na = 1 * 22.99 g = 22.99 g Cl = 1* 35.45 g = 35.45 g Total = 58.44 g 238 g NaCl * (1 mol NaCl/58.44 g NaCl) = 4.07 mol NaCl There are approximately 4.1 moles in 238 grams of sodium chloride.
(7.6 g NaCl) (moles NaCl/58.44 g NaCl) (1 mole Na/1 mole NaCl) ( 22.99 g Na/mole Na) = 2.989 g Na or 3.0 g Na (significant figures) Steps: 1. Change everything to moles: use molar mass/molecular of NaCl 2. Find the molar ratio: since there are 1 Na in every NaCl, then the ratio of NaCl to Na is 1:1 3. Change moles back to grams: use the molar mass/molecular weight of Na 4. Watch out for significant figures! I hope this is right
For this you need the atomic mass of Na. Take the number of grams and divide it by the atomic mass. Multiply by one mole for units to cancel.11.5 grams Na / (23.0 grams) = .500 moles Na
Balanced equation first. NaOH + HCl >> NaCl + H2O Everything is one to one, so 1.222 moles HCl (1mole NaOH/1mole HCl) = 1.222 moles NaOH
Every mol of NaCl contains a mol of Na, weighting 23 grams, and a mol of Cl, weighing 35.5 grams. So, every mol of NaCl weights 58,5 (=23+35.5) grams. Therefore, 145 moles of NaCl weights 8482.5 grams.
In 1 mol of NaCl there is 58.44 grams. ( 22.99 grams of Na + 35.45 grams of Cl). Using stoichiometry, you cancel the grams by taking 29.22 grams/58.44 grams. So 0.50 moles of NaCl