Question

1.How many milliliters of a 0.360 M KOH solution contain 1.75×10−2 mol of KOH?

2.Which of the following solutions will not form a precipitate when added to 0.10 M NaOH?
0.10 M NH4Br
0.10 M MgBr2
0.10 M FeCl2
3. How many grams of butanethiol can be deodorized by reaction with 4.00 mL of 9.95×10−2 M NaOCl?
Express the mass in grams to three significant digits.

Answer 1

Answer on Question #55380 – Chemistry – General Chemistry

Question:

1. How many milliliters of a 0.360 M KOH solution contain $1.75 \times 10^{-2}$ mol of KOH?

2. Which of the following solutions will not form a precipitate when added to 0.10 M NaOH?

0.10 M NH4Br, 0.10 M MgBr₂, 0.10 M FeCl₂

3. How many grams of butanethiol can be deodorized by reaction with $4.00~\mathrm{mL}$ of $9.95 \times 10^{-2}$ M NaOCl? Express the mass in grams to three significant digits.

Answer:

$$\bar{N}_M = \frac{v}{V} \quad V = \frac{v}{\bar{N}_M}$$

1)

$$V(KOH) = \frac{1.75 \cdot 10^{-2}}{0.360} = 0.05L = 48.61 \text{ ml}$$

2) NaOH+NH₄Br=NH₄OH+NaBr

$$\begin{array}{l} 2\text{NaOH} + \text{MgBr}_2 = \text{Mg(OH)}_2 + 2\text{NaBr} \\ 2\text{NaOH} + \text{FeCl}_2 = \text{Fe(OH)}_2 + 2\text{NaCl} \\ \end{array}$$

No of the formed products is insoluble compound.

All the listed solutions will not form precipitate.

3)

$$\begin{array}{l} 2C_4H_{10}S + NaOCl = C_8H_{18}S_2 + NaCl + H_2O \\ v(C_4H_{10}S) = 2 \cdot v(NaOCl) \\ v(NaOCl) = C_M(NaOCl) \cdot V(NaOCl) \\ v(NaOCl) = 0.0995 \cdot \frac{4.00}{1000} = 0.0004 \text{ mol} \\ v(C_4H_{10}S) = 2 \cdot 0.0004 = 0.0008 \text{ mol} \\ v = \frac{m}{M} \quad m = v \cdot M \\ M(C_4H_{10}S) = 90.1881 \text{ g/mol} \\ m(C_4H_{10}S) = 0.0008 \cdot 90.1881 = 0.072 \text{ g} \\ \end{array}$$

www.AssignmentExpert.com