1 mole of NaHCO3
Na = 1 * 22.99 g = 22.99 g
H = 1 * 1.01 g = 1.01 g
C = 1 * 12.01 g = 12.01 g
O = 3 * 16.00 g = 48.00 g
Total = 84.01 g
There are 84.01 grams in one mole of NaHCO3.
There are 5 atoms in one molecule of NaCO3 (Na → 1 atom + C → 1 atom + O3 → 3 atoms = 5 atoms).
In a mole (6.022 x 1023) of NaCO3 there are 3.01 x 1024 atoms.
.11905 moles
Sodium Bicarbonate = NaHCO3 Molecular weight = Summation of individual molecular weights of elements =[(1)(22.99 g/mol) ] + [(1)(1.01 g/mol)] + [(1)(12.01 g/mol)] + [(3)(15.99 g/mol)] =84.01 g/molSo,(86.6 g) / (84.01 g/mol) = 1.031 mol
Remember moles = mass(g) / Mr The 'Mr' is the relative molecular mass. The Mr(Baking soda ; sodium hydrogen carbonate ; NaHCO3) is Na x 1 = 23 x 1 = 23 H x 1 = 1 x 1 = 1 C x 1 = 12 x 1 = 12 O x 3 = 16 x 3 = 48 23 + 1 + 12 + 48 = 84 ( Mr of NaHCO3) Hence mol(NaHCO3) = 2.00g / THe answers!!!!!
3.00 grams NaHCO3 (1 mole/84.008 grams) = 0.0357 moles of sodium bicarbonate
Weighting half value of 23+1+12+(3x16) = 84/2 = 42 gram NaHCO3 gives you 0.5 mole of it, to be dissolved in about 200 mL and there after to be filled up to 500 mL (= 500 cm3) will result in a 0.5 mol / 500 mL = 0.5 mol / 0.5 L = 1 mol/L = 1M NaHCO3
1196.4423g 42.6 mol X 28.0855 g/ 1 mol = 1196.4423g
All you need is the moles and the mass of NaHCO3 and you have the moles 0.025 moles NaHCO3 (84.008 grams/1 mole NaHCO3) = 2.1 grams of sodium bicarbonate ---------------------------------------------
Divide 6.10 (g NaHCO3) by 84.007 (g.mol−1 NaHCO3) to get 0.0726 mol NaHCO3
Sodium Bicarbonate = NaHCO3 Molecular weight = Summation of individual molecular weights of elements =[(1)(22.99 g/mol) ] + [(1)(1.01 g/mol)] + [(1)(12.01 g/mol)] + [(3)(15.99 g/mol)] =84.01 g/molSo,(86.6 g) / (84.01 g/mol) = 1.031 mol
Remember moles = mass(g) / Mr The 'Mr' is the relative molecular mass. The Mr(Baking soda ; sodium hydrogen carbonate ; NaHCO3) is Na x 1 = 23 x 1 = 23 H x 1 = 1 x 1 = 1 C x 1 = 12 x 1 = 12 O x 3 = 16 x 3 = 48 23 + 1 + 12 + 48 = 84 ( Mr of NaHCO3) Hence mol(NaHCO3) = 2.00g / THe answers!!!!!
3.00 grams NaHCO3 (1 mole/84.008 grams) = 0.0357 moles of sodium bicarbonate
1 Mol
Several part problem. Get molarity of NaHCO3. (150 ml)( M NaHCO3) = (150 ml)(0.44 M HCl) = 0.44 M NaHCO3 --------------------------- get moles NaHCO3 ( 150 ml = 0.150 Liters ) 0.44 M NaHCO3 = moles NaHCO3/0.150 Liters = 0.066 moles NaHCO3 ---------------------------------------get grams 0.066 moles NaHCO3 (84.008 grams/1 mole NaHCO3) = 5.54 grams NaHCO3 needed ---------------------------------------------answer
Weighting half value of 23+1+12+(3x16) = 84/2 = 42 gram NaHCO3 gives you 0.5 mole of it, to be dissolved in about 200 mL and there after to be filled up to 500 mL (= 500 cm3) will result in a 0.5 mol / 500 mL = 0.5 mol / 0.5 L = 1 mol/L = 1M NaHCO3
1196.4423g 42.6 mol X 28.0855 g/ 1 mol = 1196.4423g
1 mole Na2CO3 = 105.98844g 0.577mol x 105.98844g/mol = 61.2g Na2CO3
8 mol x (4 mol / 2 mol) x 133.5 g / 1 mol = 2136 grams
(g) -> moles = x(g) * (1 mol/molar mass)Moles -> atoms = x(mol) * (6.022*10^23/1 mol)