The formula given shows that equal numbers of atoms of sodium and of chloride are needed to constitute one formula mass of NaCl. Therefore, 6.2 formula mass amounts of NaCl can be produced from 6.2 mol "Cl"* if at least 6.2 mol Na is also available.
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*Elemental chlorine normally occurs as diatomic molecules. However this question is answered in accordance with its own statement. 6.2 mol Cl would constitute 3.1 mol Cl2, but the answer would still be the same, because the same number of chlorine atoms would be provided for the reaction
1 mole NaCl = 58.44g 0.1601mol NaCl x 58.44g NaCl/mol NaCl = 9.35g NaCl
13g Cl2 * (1mol cl / 34.45 ) * (2 mol / 2 mol) * (58.44g NaCl / 1 mol NaCl) = 22g NaCl
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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
1 mole NaCl = 58.44g 0.1601mol NaCl x 58.44g NaCl/mol NaCl = 9.35g NaCl
I'll see if I can type it out for you.. 79g NaCl x (1 mol NaCl)/(58.44g NaCl) = 1.35 mol NaCl 1 mol NaCl = 58.44g (because Na is 22.99g and Cl is 35.45g)
13g Cl2 * (1mol cl / 34.45 ) * (2 mol / 2 mol) * (58.44g NaCl / 1 mol NaCl) = 22g NaCl
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0.0002 mol NaCl/mLmolarity = (moles solute)/(L of solution)0.20 M NaCl = 0.20 mol NaCl/L1 L = 1000 mL0.20 mole NaCl/1000 mL = 0.00020 mol NaCl/mL (rounded to two significant figures)
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
One mole of NaCl = 6,02214129(27) × 1023 mol−1 (molecules ) - this is number (constant) of Avogadro.
1 mole of NaCl is 58.44 g 200 g NaCl * (1 mol NaCl/58.44 g NaCl) = 3.422 mol NaCl There are about 3.4 moles in 200 grams of NaCl.
4Fe + 3O2 -> 2Fe2O3 is the balanced equation.It'll need enough oxygen (> (0.46*3)/4 mole O2)to give (0.46*2)/4 mole Fe2O3, so 0.23 mole Fe2O3is produced.
the molar mass of sodium is 23 and chlorine is 35 so they add together to create 58 grams per mole. So 5.58g/58g=.0962 moles.
5 grams of table salt is 5 grams of NaCl. NaCl has a molar mass of 58.443 grams/mol, so 5 grams would be .0855 mol NaCl. In one mole of NaCl there is one mole of Na, so there would be .0855 mol Na, or 5.235 * 1022 atoms Na.