What is the mass percent of carbon (12.01 g/mol) in oxalic acid, H2C2O4 (90.0349 g/mol)?
What is the mass percent of carbon (12.01 g/mol) in oxalic acid, H2C2O4 (90.0349 g/mol)?
In a standardization process, 0.2161 g of oxalic acid (H2C2O4 : MM 90.04 g/mol) was neutralized with 32.0 mL NaOH. Find the molarity of NaOH. H2C2O4 + 2NaOH à Na2C2O4 + 2H2O A: 0.250 M B: 0.100 M C: 0.150 M D: 0.300 M
What mass of oxalic acid, H2C2O4, is required to prepare 250 mL of a solution that has a concentration of (4.53x10^-1) M H2C2O4?
Data Table 1 Mass of flask and oxalic acid (g) 117.43 Mass of empty flask (g) 116.93 Mass of oxalic acid (g) 0.5 Moles of oxalic acid (mol) Final volume of NaOH (mL) 17 Initial volume of NaOH (mL) 5 Volume of NaOH used (mL) 12 Moles of NaOH (mol) Molarity of NaOH (M) Data Table 2 Mass of flask and vinegar (g) 126.61 Mass of empty flask (g) 121.63 Mass of vinegar (g) 4.98 Final volume of NaOH (mL)...
A sample of oxalic acid (a diprotic acid of the formula H2C2O4) is dissolved in enough water to make 1.00 L of solution. A 100.0 mL sample of this solution is titrated with a solution of sodium hydroxide of concentration 0.750 M and requires 20.0 mL of sodium hydroxide to reach the end point. Calculate the mass of the original oxalic acid sample.
a) Write the reaction for the neutralization oxalic acid (H2C2O4) with sodium hydroxide (NaOH). Note oxalic acid is diprotic (like sulfuric acid). b) A 0.1187 g sample of an unknown, diprotic solid acid is dissolved in water. 26.36 mL of 0.1000 M sodium hydroxide is used to reach the equivalence point. How many moles of sodium hydroxide were reacted? How many moles of acid reacted? What is the molecular weight of the acid?
A solution is prepared by dissolving 0.5892 g oxalic acid (H2C2O4) in enough water to make 100.0 mL of solution. A 10.00-mL aliquot (portion) of this solution is then diluted to a final volume of 250.0 mL. What is the final molarity of the diluted oxalic acid solution? I do not know if the answer of 0.00187 or 1.87 x 10^ -3 is correct. It is incorrect in my perspective, so i do not expect that answer, if you think...
One of the uses of oxalic acid, (H2C2O4) is rust removal. It reacts with rust, (Fe2O3) according to the equation: 2Fe2O3(s) + 6H2C2O4(aq) = 2Fe2(C2O4)3^-3(aq) + 3H2O(l) + 6H+(aq) Calculate the number of grams of rust that can be removed by 5x10^2mL of a 0.100 mol L-1 solution of oxalic acid.
Oxalic acid (H2C2O4) is a diprotic acid found in Rhubarb leaves. Write out the ionization reactions for oxalic acid when it reacts with water. Give the states involved. step 1: step 2:
6. Oxalic acid H2C2O4 is a diprotic acid and can be used, in dilute form, as wood bleach. pKa1 = 1.237 and pKa2=4.187 A. How many milliliters of 0.0500 M NaOH would be required to completely neutralize 10.0 mL of 0.100 M H2C2O4? (5 pts) B. Estimate the pH at the first equivalence point of the titration. (6 pts) C. What is the pH at the second equivalence point of the titration. (8 pts) Please help with work shown, especially...
one practical application of oxalic acid (H2C2O4) is as a rust remover. Rust is primarily composed of Fe2O3, which reacts with oxalic acid according to the unbalanced equation shown below. Fe2O3 (s) + H2C2O3 (aq) --> Fe(C22O4)3-3 (aq) + H2O (l) + H+ (aq) Calculate the number of milligrams of Fe2O3 that can be removed using 500.0 mL of 0.1068 M H2C2O4.