0.5054 grams of soda ash was diluted with 25 mL of water. Itwas titrated with 90

Question

0.5054 grams of soda ash was diluted with 25 mL of water. Itwas titrated with 90.51 mL of 0.08042M HCl. Calculate the moles and hence mass of sodium carbonate in thesoda ash. 0.5054 grams of soda ash was diluted with 25 mL of water. Itwas titrated with 90.51 mL of 0.08042M HCl. Calculate the moles and hence mass of sodium carbonate in thesoda ash.

Explanation / Answer

Na2CO3 + 2HCl -> 2NaCl + H2O +CO2 n(HCl) = cv = 0.08042 x 0.09051 = 0.007279mol n(Na2CO3) = 1/2 x 0.007279 = 0.0036394mol m(Na2CO3) = nM = 0.0036394 x 105.99 =0.386g Hope this helps!

Related Questions

Navigate

• Browse (All)
• Browse
• Subjects

---

---

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