Phần trăm khối lượng nguyên tử cacbon có trong phân tử Salbutamol là

Ta có: =12.13+1.21+14+16.312.13​.100%≈65,272% %��=12.1312.13+1.21+14+16.3.100%≈65,272





 

$Mn{O}^2+4HCl\underrightarrow{t^o}MnC{l}^2+C{l}^2+2H^{2}O$

$C{l}^2+H^2\underrightarrow{as}2HCl$

$NaOH+HCl\to NaCl+H^{2}O$

$2NaCl+2H^{2}O\underrightarrow{đpcmn}2NaOH+H^2+C{l}^2$
 

MgCO3 + 2HCl --> MgCl2 + H2O + CO2;

0,1----------0,2---------0,1---------------------0,1 (mol)

MgO + 2 HCl ---> MgCl2 + H2O;

0,05--------0,1-------------0,05 (mol)

a. Ta có: nCO2= 2,24/22,4=0,1 (mol);=> nMgCO3= 0,1 (mol);

=> mMgCO3= 0,1*84=8,4(g);=> mMgO=10,4-8,4=2(g).

b.nMgO= 2/40=0,05(mol)

Theo pthh: nHCl cần= 0,2+0,1=0,3(mol);=> mHCl=0,3*36,5= 10,95(g)

=>mddHCl=$\frac{10,95\cdot 100}{7,3}=150\left(g\right)$

c.nMgCl2 sr= 0,1+0,05=0,15 (mol);=> mMgCl2=0,15*95=14,25(g)

mCO2 thoát ra= 0,1*44=4,4(g)

mdd sau pư= 10,4+150-4,4=156(g)

C% MgCl2=$\frac{14,25\cdot 100}{156}=9,13\%$

a, PT: $2C^2{H}^2+5O^2\underrightarrow{t^o}4C{O}^2+2H^{2}O$

Ta có: $n^{C^2{H}^2}=\genfrac{}{}{0ex}{}{13}{26}=0,5\left(mol\right)$

Theo PT: $n^{C{O}^2}=n^{C^2{H}^2}=1\left(mol\right)\Rightarrow m^{C{O}^2}=1.44=44\left(g\right)$

$n^{H^{2}O}=n^{C^2{H}^2}=0,5\left(mol\right)\Rightarrow m^{H^{2}O}=0,5.18=9\left(g\right)$

b, Theo PT: $n^{O^2}=\genfrac{}{}{0ex}{}{5}{2}{n}^{C^2{H}^2}=1,25\left(mol\right)\Rightarrow V^{O^2}=1,25.22,4=28\left(l\right)$

$\Rightarrow V^{kk}=\genfrac{}{}{0ex}{}{V^{O^2}}{20\%}=140\left(l\right)$

c, PT: $C{O}^2+Ca{\left(OH\right)}^2\to CaC{O}^3+H^{2}O$

Theo PT: $n^{CaC{O}^3}=n^{C{O}^2}=1\left(mol\right)\Rightarrow m^{CaC{O}^3}=1.100=100\left(g\right)$