PTHH: \(Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\\ 0,1mol:0,2mol\rightarrow0,2mol:0,1mol:0,1mol\)
\(n_{Na_2CO_3}=\dfrac{10,6}{106}=0,1\left(mol\right)\)
a. \(n_{NaCl}=0,2.58,5=11,7\left(g\right)\)
b. \(V_{CO_2}=0,1.22,4=2,24\left(l\right)\)
c. \(m_{HCl}=0,2.36,5=7,3\left(g\right)\)
\(n_{Na_2CO_3}=\dfrac{m_{Na_2CO_3}}{M_{Na_2CO_3}}=\dfrac{10,6}{106}=0,1\left(mol\right)\)
PTHH:
\(Na_2CO_3+2HCl\rightarrow2NaCl+H_2O+CO_2\)
\(0,1............0,2........0,2.......0,1.......0,1\) (mol)
a. \(m_{NaCl}=M_{NaCl}.n_{NaCl}=0,2.58,5=11,7\left(g\right)\)
b. \(V_{CO_2}=22,4.n_{CO_2}=22,4.0,1=2,24\left(l\right)\)
c. \(m_{HCl}=n_{HCl}.M_{HCl}=0,2.36,5=7,3\left(g\right)\)
\(n_{Na_2CO_3}=\dfrac{10,6}{106}=0,1\left(mol\right)\)
PTHH: \(Na_2CO_3+2HCl\rightarrow2NaCl+H_2O+CO_2\)
a. Theo PT ta có: \(n_{NaCl}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)
\(\Rightarrow m_{NaCl}=0,2.58,5=11,7\left(g\right)\)
b. Theo PT ta có: \(n_{CO_2}=n_{Na_2CO_3}=0,1\left(mol\right)\)
\(\Rightarrow V_{CO_2}=0,1.22,4=2,24\left(l\right)\)
c. Theo PT ta có: \(n_{HCl}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,2.36,5=7,3\left(g\right)\)
Na2CO3+2HCl→2NaCl+H2O+CO2
a) nNa2CO3= mNa2CO3:MNa2CO3=10,6:106=0,1(mol)
nNaCl=0,2.1:1=0,2(mol)
mNaCl=nNaCl.MNaCl= 0,2.58,5=11,7(g)
b) VCo2=nCo2.22,4=0,1.1:1.22,4=2,24(lít)
c) mHCl=nHCl.MHCl= 0,1.2:1:36,5=7,3(g)
