$a,m_{H_2SO_4}=50.9,8\%=4,9g.$

$⇒m_{H_2O}=50-4,9=45,1g.$

$b,PTPƯ:Zn+H_2SO_4\xrightarrow{} ZnSO_4+H_2↑$

$n_{H_2SO_4}=\dfrac{4,9}{98}=0,05mol.$

$Theo$ $pt:$ $n_{H_2}=n_{H_2SO_4}=0,05mol.$

$⇒V_{H_2}=0,05.22,4=1,12l.$

$c,m_{H_2SO_4}=4,9+(150.8\%)=16,9g.$

$m_{ddH_2SO_4}=50+150=200g.$

$⇒C\%_{H_2SO_4}=\dfrac{16,9}{200}.100\%=8,45\%$

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Đáp án:

 a) 4,9g vaf 45,1 g

b) 1,12l

c) 8,45%

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
m{H_2}S{O_4} = mdd \times C\%  = 50 \times 9,8\%  = 4,9g\\
m{H_2}O = mdd - m{H_2}S{O_4} = 50 - 4,9 = 45,1g\\
b)\\
Zn + {H_2}S{O_4} \to ZnS{O_4} + {H_2}\\
n{H_2} = n{H_2}S{O_4} = \dfrac{m}{M} = \dfrac{{4,9}}{{98}} = 0,05\,mol\\
V{H_2} = n \times 22,4 = 0,05 \times 22,4 = 1,12l\\
c)\\
m{H_2}S{O_4} = 150 \times 8\%  = 12g\\
C\%  = \dfrac{{12 + 4,9}}{{50 + 150}} \times 100\%  = 8,45\% 
\end{array}\)