Hóa học

Trích mẫu thử và đánh số thứ tự từng chất:

C2H2 + AgNO3 -----------> 2Ag + 2CO2 + H2O 

\(m_{NaOH}=\dfrac{20\cdot20\%}{100\%}=4\left(g\right)\)

\(n_{NaOH}=\dfrac{4}{40}=0,1\left(mol\right)\)

PTHH:    \(2NaOH+CuSO_4\rightarrow Na_2SO_4+Cu\left(OH\right)_2\downarrow\)

Mol (pt):      2                  1                   1                   1 

Mol (đề):    0,1

a) Theo pt, ta có: \(n_{CuSO_4}=2\cdot n_{NaOH}=2\cdot0,1=0,2\left(mol\right)\)

\(\Rightarrow m_{CuSO_4}=0,2\cdot160=32\left(g\right)\)

\(\Rightarrow m_{\text{dd }CuSO_4}=\dfrac{32\cdot100\%}{10\%}=320\left(g\right)\)

b) Theo pt, ta có: \(n_{Cu\left(OH\right)_2}=n_{Na_2SO_4}=n_{CuSO_4}=0,2\left(mol\right)\)

\(\Rightarrow m_{Cu\left(OH\right)_2}=0,2\cdot98=19,6\left(g\right)\)

c) \(m_{Na_2SO_4}=0,2\cdot142=28,4\left(g\right)\)

Theo ĐLBTKL, ta có:

\(m_{\text{dd }Na_2SO_4}=m_{\text{dd }NaOH}+m_{\text{dd }CuSO_4}-m_{Cu\left(OH\right)_2}\)

\(=20+320-19,6=320,4\left(g\right)\)

\(\Rightarrow C\%_{\text{dd }Na_2SO_4}=\dfrac{28,4}{320,4}\cdot100\%\approx8,86\%\)

\(\text{#}Toru\)

bạn xem lại đề

\(n_{CO_2}=\dfrac{4}{44}=\dfrac{1}{11}mol\\ CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\\ n_{CaCO_3}=n_{CO_2}=\dfrac{1}{11}mol\\ m_{CaCO_3}=\dfrac{1}{11}\cdot100=\dfrac{100}{11}g\\ \%CaCO_3=\dfrac{100:11}{10}\cdot100\%=90,91\%\)

\(V_{C_2H_5OH}=\dfrac{36,8.90}{100}=33,12ml\\ m_{C_2H_5OH}=33,12.0,8=26,496g\\ n_{C_2H_5OH}=\dfrac{26,496}{46}=0,576mol\\ 2C_2H_5OH+2K\rightarrow2C_2H_5OK+H_2\\ n_{H_2}=\dfrac{1}{2}n_{C_2H_5OH}=0,288mol\\ V_{H_2}=0,228.22,4=6,4512l\)

Câu 2:

\(2Na+2H_2O\rightarrow2NaOH+H_2\)

\(2Na+2C_2H_5OH\rightarrow2C_2H_5ONa+H_2\uparrow\)

Câu 1: 

\(V_{C_2H_5OH}=\dfrac{40\cdot96}{100}=38,4\left(ml\right)\)

\(m_{C_2H_5OH}=38,4\cdot0,8=30,72\left(gam\right)\)

\(n_{C_2H_5OH}=\dfrac{30,72}{12\cdot2+6+16}=\dfrac{384}{575}\left(mol\right)\)

=>\(n_{H_2}=\dfrac{192}{575}\left(mol\right)\)

\(V_{H_2}=\dfrac{192}{575}\cdot22,4\simeq7,48\left(lít\right)\)