Question

1. How many moles of BaCl2 are formed in the neutralization of 196.5 mL of 0.095 M Ba(OH)2 with aqueous HCl?
2. Lead ions an be precipitated from aqueous solutions by the addition of aqueous iodide: Pb+2(aq) + 2I-2 (aq) =PbI2(s) . Lead iodideis virtually insoluble in water so that the reaction appears to go to completion. How many milliliters of 1.180 M HI(aq) must be added to a solution containing 0.200 mol of Pb(NO3)2 (aq) to completely precipitate the lead?
3. What is the molarityof a NaOH solution if 15.5 mL of a 0.220 M H2SO4 solution is required to neutralize a 25.0 mL sample of the NaOH solution?
4. Aqueous solutions of a compound did not form precipitates with Cl-, Br-, I-, SO4-2, CO3-2, PO4-3, OH-, or S-2. This highly water soluble compound produced the foul-smelling gas H2S when the solution was acidified. This compound is _______.

Answer 1

Answer on Question #55401 - Chemistry - General chemistry

Question:

1. How many moles of BaCl₂ are formed in the neutralization of 196.5 mL of 0.095 M Ba(OH)₂ with aqueous HCl?

2. Lead ions can be precipitated from aqueous solutions by the addition of aqueous iodide: Pb+2(aq) + 2I-2 (aq) = PbI2(s). Lead iodide is virtually insoluble in water so that the reaction appears to go to completion. How many milliliters of 1.180 M Hl(aq) must be added to a solution containing 0.200 mol of Pb(NO₃)₂ (aq) to completely precipitate the lead?

3. What is the molarity of a NaOH solution if 15.5 mL of a 0.220 M H₂SO₄ solution is required to neutralize a 25.0 mL sample of the NaOH solution?

4. Aqueous solutions of a compound did not form precipitates with Cl-, Br-, I-, SO₄²⁻, CO₃²⁻, PO₄³⁻, OH⁻, or S-2. This highly water soluble compound produced the foul-smelling gas H₂S when the solution was acidified. This compound is ________.

Solution:

1. $Ba(OH)_2 + 2HCl \rightarrow BaCl_2 + 2H_2O$

$$C = \frac{n}{V}$$ $$n = V \times C$$ $$n = 1.965 \times 10^{-2} \times 0.095$$ $$n = 1.87 \times 10^{-2}$$ $$n_{(Ba(OH)2)} = n_{BaCl2}$$

Answer: $1.87 \times 10^{-2}$ moles

2. $Pb(NO_3)_2 + 2HI \rightarrow PbI_2 + 2HNO_3$

$$C = \frac{n}{V}$$ $$V = \frac{n}{C}$$ $$n(HI) = 2 \times n(Pb(NO_3)_2) = 0.4 \text{ mol}$$ $$V = \frac{0.4}{1.18}$$

Answer: $V = 0.34 \text{ ml}$

3. $2NaOH + H_{2}SO_{4}\rightarrow Na_{2}SO_{4} + H_{2}O$

$$n = V \times C$$ $$n = 0.0155 \times 0.220$$ $$n = 0.0341$$ $$n(NaOH) = 0.0341 \times 2$$ $$n(NaOH) = 0.0682$$ $$C = \frac{n}{V}$$ $$C = \frac{0.0682}{0.025}$$ $$C = 0.27$$

Answer: 0.27 mol/L

4. $(NH_4)_2S$

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