First determine the mass of 2 lbs of NaOH - it's about 2 x 454 g = 908 g
[Having to use a conversion factor introduces error.]
A mole of NaOH is 40 g; 908 g/40 g/mole = 22.7 moles.
2 pounds = 907.18474 grams
Since both the acid and the base have equivalent weights equal to their formula weights, 2 moles of KOH are needed to neutralize 2 moles of nitric acid.
4700
2 pounds 9 ounces is 41 ounces.
907.2 grams per two pounds.
Need moles sodium hydroxide first.80 grams NaOH (1 mole NaOH/39.998 grams) = 2.000 moles NaOHMolarity = moles of solute/Liters of solutionMolality = 2.000 moles NaOH/1 liter= 2 M NaOH solution===============
To find the number of moles in 80.0g of NaOH, divide the given mass by the molar mass of NaOH. The molar mass of NaOH is approximately 40.0 g/mol (sodium - 22.99 g/mol, oxygen - 16.00 g/mol, hydrogen - 1.01 g/mol). Therefore, there are 2.00 moles of NaOH in 80.0g.
This ratio is 1:2.
This reaction is:2 Al + 2 H2O + 2 NaOH = 2 NaAlO2 + 3 H2From 4 moles of Al 6 moles of hydrogen are obtained.
4 moles or 160 g NaOH is required for one litre solution.
Write out the equation, and remember to balance each side.Na2CO3 + Ca(OH)2 --> 2NaOH + CaCO3Molecular WeightsNa2CO3: 106 grams/moleNaOH: 40 grams/moleAlways convert your reagents into moles.(120g Na2CO3) x (1 mole Na2CO3/106 grams Na2CO3) = 1.132 molesAccording to the balanced equation, 1 molecule of Na2CO3 generates 2 molecules of NaOH.(1.132 moles Na2CO3) x (2 moles NaOH/1 mole Na2CO3) = 2.264 moles NaOHNow determine the number of grams from 2.264 moles of NaOH.(2.264 moles NaOH) x (40 grams/ 1 mole NaOH) = 90.57 grams NaOH formed.To prevent rounding off too many times, carry out the dimensional analysis in one step:(120g Na2CO3) x (1 mole Na2CO3/106 grams Na2CO3) x(2 moles NaOH/1 mole Na2CO3) x (40 grams/ 1 mole NaOH) = 90.57 grams NaOH
2 moles of NaOH will react with 1 mole of H2SO4 based on the balanced chemical equation: 2NaOH + H2SO4 -> Na2SO4 + 2H2O.
The molarity of the solution can be calculated by dividing the moles of solute by the volume of solution in liters. In this case, 2 moles of NaOH in 1620 mL (1.62 L) of water gives a molarity of approximately 1.23 M.
Well, darling, if we're talking about a 1:2 molar ratio between NaOH and H2SO4, then you'd need 2 moles of NaOH to neutralize 1 mole of H2SO4. It's all about those stoichiometry dance moves, honey. Just make sure you're not tripping over your chemical equations!
Since H2SO4 is a diprotic acid, it will require twice the amount of NaOH to neutralize it. Therefore, molarity of NaOH should also be 1 M. 1 mole of H2SO4 reacts with 2 moles of NaOH. Therefore, to neutralize 1 mole of H2SO4, 2 moles of NaOH are required. To neutralize 1 mole of H2SO4 in 100 ml (0.1 L) of 1 M solution, you will need 0.1 moles of NaOH.
Approximately 4.50 grams of NaOH are required to prepare 200 mL of a 0.450 M solution. This can be calculated using the formula: moles = Molarity x Volume (in L), and then converting moles to grams using the molar mass of NaOH.
2 m