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Chapter 17: Problem 63

A quantity of \(0.560 \mathrm{~g}\) of \(\mathrm{KOH}\) is added to \(25.0 \mathrm{~mL}\) of \(1.00 \mathrm{M} \mathrm{HCl}\). Excess \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) is then added to the solution. What mass (in grams) of \(\mathrm{CO}_{2}\) is formed?

Short Answer

Expert verified
The mass of \(\mathrm{CO}_2\) that is formed is approximately \(0.660\) g.

Step by step solution

01

Calculate moles of KOH

First, calculate the moles of \(\mathrm{KOH}\). The molar mass of \(\mathrm{KOH}\) is approximately \(56.11 \mathrm{~g/mol}\). So the moles of \(\mathrm{KOH}\) are \(\frac{0.560 ~g}{56.11 ~g/mol} \approx 0.01 \mathrm{~mol}\) .
02

Calculate moles of HCl initially and after reaction with KOH

Initially, \(25.0\) \(\mathrm{mL}\) of \(1.00 \) \(\mathrm{M} \mathrm{HCl}\) corresponds to \(1.00 \mathrm{M} * 0.025 ~L = 0.025 \mathrm{~mol}\). All the \(\mathrm{KOH}\) reacts with the \(\mathrm{HCl}\), using up \(0.01 \mathrm{~mol}\) of \(\mathrm{HCl}\). So the moles of \(\mathrm{HCl}\) remaining are \(0.025 \mathrm{~mol} - 0.01 \mathrm{~mol} = 0.015 \mathrm{~mol}\).
03

Calculate moles of CO2 formed

Each mole of \( \mathrm{HCl} \) reacts with \( \mathrm{Na}_2 \mathrm{CO}_3 \) to produce one mole of \( \mathrm{CO}_2 \). So, \( 0.015 \)moles of \( \mathrm{HCl} \) will produce \( 0.015 \) moles of \( \mathrm{CO}_2 \) .
04

Calculate mass of CO2 formed

Finally, convert moles of \( \mathrm{CO}_2 \) to grams using its molar mass (around \(44.01 \)g/mol): \(0.015 \)mol * \(44.01 \)g/mol = 0.660 g.

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Most popular questions from this chapter

The molar solubility of \(\mathrm{AgCl}\) in \(6.5 \times 10^{-3} \mathrm{M}\) \(\mathrm{AgNO}_{3}\) is \(2.5 \times 10^{-8} M .\) In deriving \(K_{\mathrm{sp}}\) from these data, which of these assumptions are reasonable? (a) \(K_{\mathrm{sp}}\) is the same as solubility. (b) \(K_{\mathrm{sp}}\) of \(\mathrm{AgCl}\) is the same in \(6.5 \times 10^{-3} \mathrm{M} \mathrm{AgNO}_{3}\) as in pure water. (c) Solubility of \(\mathrm{AgCl}\) is independent of the concentration of \(\mathrm{AgNO}_{3}\) (d) \(\left[\mathrm{Ag}^{+}\right]\) in solution does not change significantly on the addition of \(\mathrm{AgCl}\) to \(6.5 \times 10^{-3} \mathrm{M} \mathrm{AgNO}_{3}\). (e) \(\left[\mathrm{Ag}^{+}\right]\) in solution after the addition of \(\mathrm{AgCl}\) to \(6.5 \times 10^{-3} \mathrm{MAgNO}_{3}\) is the same as it would be in pure water.

A student wishes to prepare a buffer solution at \(\mathrm{pH}=\) \(8.60 .\) Which of these weak acids should she choose and why: HA \(\left(K_{\mathrm{a}}=2.7 \times 10^{-3}\right),\) HB \(\left(K_{\mathrm{a}}=4.4 \times\right.\) \(10^{-6}\) ), or HC \(\left(K_{\mathrm{a}}=2.6 \times 10^{-9}\right) ?\)

Calculate the \(\mathrm{pH}\) of \(1.00 \mathrm{~L}\) of the buffer \(1.00 \mathrm{M}\) \(\mathrm{CH}_{3} \mathrm{COONa} / 1.00 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}\) before and after the addition of (a) \(0.080 \mathrm{~mol} \mathrm{NaOH}\) and (b) \(0.12 \mathrm{~mol}\) HCl. (Assume that there is no change in volume.)

Solid NaI is slowly added to a solution that is \(0.010 M\) in \(\mathrm{Cu}^{+}\) and \(0.010 \mathrm{M}\) in \(\mathrm{Ag}^{+}\). (a) Which compound will begin to precipitate first? (b) Calculate \(\left[\mathrm{Ag}^{+}\right]\) when CuI just begins to precipitate. (c) What percentage of \(\mathrm{Ag}^{+}\) remains in solution at this point?

The \(\mathrm{p} K_{\mathrm{a}}\) of butyric acid (HBut) is 4.7 . Calculate \(K_{\mathrm{b}}\) for the butyrate ion (But \(^{-}\) ).
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