How many g of succinic acid (H.c4H404, 118.09 g/mol) are required to completely

Question

How many g of succinic acid (H.c4H404, 118.09 g/mol) are required to completely 8. react with (neutralize) 25.00 mLofa 0.1039 M solution of sodium hydroxide (NaOH)? H,C4H404 2 NaOH 2 H2O Na C4H404 (a) 1.100 x 102 g (b) 5.195 x 10 g (c) 0.1534 g (d) 0.3067 g (e) 0.6135 g a 20. CHM mols a determined the concentration of an HC In lab, a 11500 student experimentally solution to be 0.1875 M. The actual concentration was 0.1762 M. Calculate the error in this result. (a) 1.13% (b) 6.03% 0,064 (c) 6.41% 0 1762 (d) 94.0% (e) 98.9%

Explanation / Answer

Given

Molarity of NaOH = 0.1039 M = 0.1039 mol/L

Volume of NaOH = 25 ml = 0.025 L

No. of moles of NaOH = Volume * Molarity = 0.1039 mol/L * 0.025 L = 2.5975 * 10-3 moles of NaOH

according given reaction stoichometry

2 moles of NaOH will react with 1 moles of succinic acid

so 2.5975 * 10-3 moles of NaOH will react with (1/2) * 2.5975 * 10-3 = 1.29875 * 10-3 moles of succinic acid

Molar mass of succinic acid = 118 g/mol

Mass of succinic acid = Molar mass * no. of moles = 1.29875 * 10-3 moles * 118 g/mol = 0.1534 g

Answer is (c) 0.1534 g

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.