This is the carbon dioxide.
The limiting reagent in a reaction is the reactant that runs out first. For example, if you are reacting 10 moles of HCl and 5 moles of NaOH, you will get 5 moles of H20, 5 moles of NaCl, and 5 moles of HCl, because the remaining HCl had nothing to react with. Therefore, the NaOH is the limiting reagent.
The chemical equation is:2 NaOH + H2SO4 = Na2SO4 + 2 H2OMolar mass of sodium hydroxide is 39,9971 g; molar mass of sulfuric acid is 98,079 g.2 . 39,9971 g NaOH----------------------98,079 g H2SO4200 g NaOH------------------------xx = (200 x 98,079)/2 . 39,9971 = 245 g H2SO4So sulfuric acid is the limiting reagent.
card. + Kedde's reagent (3,5- dinitrobenzoic acid + NaOH → violet colour.
The mass of lead(II) nitrate required to react with 370 g NaOH is 1 531,9 g.
Phenol is not dissolved in a sodium hydroxide solution; having the characteristics of a weak acid phenol react with NaOH.
The limiting reagent in a reaction is the reactant that runs out first. For example, if you are reacting 10 moles of HCl and 5 moles of NaOH, you will get 5 moles of H20, 5 moles of NaCl, and 5 moles of HCl, because the remaining HCl had nothing to react with. Therefore, the NaOH is the limiting reagent.
Sodium hydroxide is the limiting reagent.
vanillin violently reacts with Bromine in carbon tetrachloride,tollens reagent and aqueous NaOH
The limiting reactant or reagent can be determined by calculating the number of moles of each reactant/reagent. Whichever is the lowest number of moles is the limiting reagent in the reaction, assuming that stoichiometry is 1;1
The chemical equation is:2 NaOH + H2SO4 = Na2SO4 + 2 H2OMolar mass of sodium hydroxide is 39,9971 g; molar mass of sulfuric acid is 98,079 g.2 . 39,9971 g NaOH----------------------98,079 g H2SO4200 g NaOH------------------------xx = (200 x 98,079)/2 . 39,9971 = 245 g H2SO4So sulfuric acid is the limiting reagent.
Alcohols react with NaOH, as ethanol forms sodium ethoxide with NaOH.
card. + Kedde's reagent (3,5- dinitrobenzoic acid + NaOH → violet colour.
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
5 mL of NaOH
green
Methoxyphenol is a phenol derivative and so, is soluble in water. This means that methoxyphenol will dissolve in an aqueous NaOH solution, but will not react with the NaOH.
Any reaction occur.