1 mol Na X (22.99 Na / mol Na) = 22.99 g Na1 mol N X (14.01 g N / mol N) = 14.01 g N3 mol O X (16.00 g O / mol O) = 48.00 g OMolar mass of NaNO3 = 85.00 g/mol
Molar mass Na = 22.99g/mol0.15 mol Na = 0.15*22.99 = 3.45g
First write balance equation 2Na + 2H2O --> 2NaOH + H2 molar mass Na = 22.99 g/mol molar mass H2O = 18.02 g/mol molar mass NaOH = 40.0 g/mol Determine limiting reagent. 1.20 g Na * 1mol Na/22.99g Na = 0.05219 (0.0522) mol Na Since 2Na = 2NaOH in balanced equation, The mol of NaOH is also 0.0522 3ml h20 = 3 gram h20 (1ml^3=1g^3) 3g H20 * 1mol/18gH20 = 0.167 (0.17) mol H20 So .17 ml H20 equals .17 mol NaOH and .0522 mole Na equals .0522 mol NaOH The smaller one is the limiting reagent, which in this case is Na 0.0522mol NaOH*40g NaOH/1mol NaOH = 2.09 gram NaOH
All you really need here is a ratio of molar masses, ie molar mass of Na/molar mass of NaBr So look at your periodic table and is says: Na = 23g/mol Br = 80g/mol therefore NaBr = 103g/mol so from this, you actually know a percentage of Sodium to Bromine, correct me if I'm wrong, someone else please but just take: (23g/mol)/103g/mol = 22.3% of you NaBr is actually Na, so take 5.35g * 0.223 and you get: 1.19g of Na in your NaBr Hope that helps. Cheers
An Na(+) ion is just one electron short of a neutral atom. Electrons have a mass of 9.109 X 10^-31 Kilograms. Let us ignore that. 0.187 mol Na(+) (6.022 X 10^23/1mol Na(+) = 1.13 X 10^22 ions of Na(+)
molar mass Na = 22.99g/mol (atomic weight in grams) 0.0135mol Na x (22.99g/mol) = 0.310g Na
1 mol Na X (22.99 Na / mol Na) = 22.99 g Na1 mol N X (14.01 g N / mol N) = 14.01 g N3 mol O X (16.00 g O / mol O) = 48.00 g OMolar mass of NaNO3 = 85.00 g/mol
1 mol Na X (22.99 Na / mol Na) = 22.99 g Na1 mol N X (14.01 g N / mol N) = 14.01 g N3 mol O X (16.00 g O / mol O) = 48.00 g OMolar mass of NaNO3 = 85.00 g/mol
1 mole Na = 22.989770g 9 mol Na x 22.989770g/mol = 206.9g Na = 200g Na rounded to one significant figure
Use this fomula: (m/M)*NA in which m is mass in gram (g), M is molar mass (g/mol) and NA is the Avogadro number (mol^-1)
Molar mass Na = 22.99g/mol0.15 mol Na = 0.15*22.99 = 3.45g
Molar mass of Na = 22.9898g/mol Molar mass of Cl = 35.45g/mol Total molar mass = 58.4398g/mol 58.4398g * 3.7mol =216.22726g
Molar mass Na: 23 g/mol O : 16 g/mol H : 1 g/mol Total NaOH: 40 g/mol Mass percentage %Na: 23/40 = 57.5% %O : 16/40 = 40% %H : 1 /40 = 2.5%
1 mol of Na+ = 22.989g Na+(0.350 mol) x (22.989 g/mol) = 8.05 g Na+
22.9 + 35.3 + 15.9 = 74.1 g/mol Na: 22.9 g/mol Cl: 35.3 g/mol O: 15.9 g/mol
Sodium (Na) has atomic number 11 and Atomic Mass of 22.99 g/mol
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.