Question

A 0.17g sample of a Group 14 chloride, XCl 4, reacted with water to produce an oxide, XO2, and
HCl.
equation 1 XCl 4(s) + 2H2O(l) → XO2(s) + 4HCl (aq)
The HCl produced was absorbed in 100 cm3
of 0.10 mol dm–3 sodium hydroxide solution (an excess).
In a titration, the unreacted sodium hydroxide solution required 30.0 cm3
of 0.20 mol dm–3 hydrochloric
acid for complete neutralisation.
(a) Calculate the amount, in moles, of hydrochloric acid used in the titration to neutralise the
unreacted sodium hydroxide solution.
amount = ............................ mol

Answer 1

Answer on Question #77577, Chemistry / Physical Chemistry

Question:

A 0.17 g sample of a Group 4 chloride, XCl₄, reacted with water to produce an oxide, XO₂, and HCl.

Equation:

$$1 \times \text{Cl}_4(\text{s}) + 2 \text{H}_2\text{O}(\text{l}) \rightarrow \text{XO}_2(\text{s}) + 4 \text{HCl}(\text{aq})$$

The HCl produced was absorbed in 100 cm³ of 0.10 mol/dm³ sodium hydroxide solution (an excess). In a titration, the unreacted sodium hydroxide solution required 30.0 cm³ of 0.20 mol/dm³ hydrochloric acid for complete neutralization.

(a) Calculate the amount, in moles, of hydrochloric acid used in the titration to neutralize the unreacted sodium hydroxide solution.

amount = …………………… mol

Solution:

Amount of HCl used in the titration:

$$0.20 \cdot 0.0300 = 0.006 \text{ mol}$$

I suppose the full solution should be as follows:

Amount of unreacted NaOH: 0.006 mol

Starting amount of NaOH:

$$0.10 \cdot 0.100 = 0.01 \text{ mol}$$

Amount of reacted NaOH:

$$0.01 - 0.006 = 0.004 \text{ mol}$$

Amount of produced HCl: 0.004 mol

Amount of XCl₄:

$$0.004 / 4 = 0.001 \text{ mol}$$

Molar Weight of XCl₄:

$$0.17 / 0.001 = 170 \text{ g/mol}$$

Atomic weight of X:

$$170 - (4 \cdot 35.45) = 28.20 \text{ g/mol}$$

X is Silicon (Si)

XCl₄ is SiCl₄

Answer:

(a) amount = 0.006 mol (also see solution above)

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