How many grams of lithium bromide would be present after evaporation when 250.0

Question

How many grams of lithium bromide would be present after evaporation when 250.0 mL of 1.00 M HBr and 250 0 ml of 1.00 M LiOH are mixed together? Is either HBr or LiOH a limiting reagent? When 5.43 g of oxalic acid (H_2C_2O_4) is placed in water containing phenol red, the suspension is yellow. A solution of KOH is added until the color changes to red. This color change is obtained at the equivalence point of this titration. What is the molarity of KOH if 32.2 mL is required to reach the equivalence point? A 1.20 g tablet of Mg(OH)_2 neutralizes 325.0 mL of stomach acid What is the molairty of the HCI in the solution? Assume stomach acid is the HCl solution.

Explanation / Answer

(10)

HBr (aq.) + NaOH (aq.) --------------> NaBr (aq.) + H2O (l)

Moles of HBr = 1.00 * 250.0 / 1000 = 0.250 mol

Moles of NaOH = 1.00 * 250.0 /1000 = 0.250 mol

SO, both are in equal quatity and both are strong electrolytes, so no one is left after reaction.

No one is limiting reagent.

And from the above balanced equation,

1 mol of acid or base form 1 mol of lithium bromide

then, 0.250 mol of acid or base form 0.250 mol lithium bromide.

Now, the mass lithium bromide fromed = no.of moles * molar mass = 0.250 * 86.845 = 21.7 g.

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

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