Đáp án:
\(\begin{array}{l}
b)\\
{m_{Cu{{(OH)}_2}}} = 14,7g\\
c)\\
{C_{{M_{N{a_2}S{O_4}}}}} = 1,5M\\
{C_{{M_{NaOH}}}} = 1M\\
d)\\
C{\% _{N{a_2}S{O_4}}} = 17,56\% \\
C{\% _{NaOH}} = 3,3\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2NaOH + CuS{O_4} \to N{a_2}S{O_4} + Cu{(OH)_2}\\
b)\\
{n_{NaOH}} = \dfrac{m}{M} = \dfrac{{16}}{{40}} = 0,4mol\\
{n_{CuS{O_4}}} = \dfrac{m}{M} = \dfrac{{24}}{{160}} = 0,15mol\\
\dfrac{{0,4}}{2} > \dfrac{{0,15}}{1} \Rightarrow NaOH\text{ dư}\\
{n_{Cu{{(OH)}_2}}} = {n_{CuS{O_4}}} = 0,15mol\\
{m_{Cu{{(OH)}_2}}} = n \times M = 0,15 \times 98 = 14,7g\\
c)\\
{n_{N{a_2}S{O_4}}} = {n_{CuS{O_4}}} = 0,15mol\\
{C_{{M_{N{a_2}S{O_4}}}}} = \dfrac{n}{V} = \dfrac{{0,15}}{{0,1}} = 1,5M\\
{n_{NaO{H_d}}} = {n_{NaOH}} - 2{n_{CuS{O_4}}} = 0,1mol\\
{C_{{M_{NaOH}}}} = \dfrac{n}{V} = \dfrac{{0,1}}{{0,1}} = 1M\\
d)\\
{m_{{\rm{dd}}NaOH}} = 60 \times 1,2 = 72g\\
{m_{{\rm{dd}}CuS{O_4}}} = 40 \times 1,6 = 64g\\
{m_{{\rm{dd}}spu}} = 72 + 64 - 14,7 = 121,3g\\
{m_{N{a_2}S{O_4}}} = n \times M = 0,15 \times 142 = 21,3g\\
{m_{NaO{H_d}}} = n \times M = 0,1 \times 40 = 4g\\
C{\% _{N{a_2}S{O_4}}} = \dfrac{{21,3}}{{121,3}} \times 100\% = 17,56\% \\
C{\% _{NaOH}} = \dfrac{4}{{121,3}} \times 100\% = 3,3\%
\end{array}\)
