\(\text{Na2CO3+2HCl->2NaCl+CO2+H2O}\)
a, Ta có :
\(n_{Na2CO3}=\frac{10,6}{106}=0,1\left(mol\right)\)
\(n_{HCl}=0,2.2=0,4\left(mol\right)\)
=>HCl dư
Vậy quỳ tím có màu đỏ
b)Ta có :
\(n_{CO2}=n_{Na2CO3}=0,1\left(mol\right)\)
\(\Rightarrow V_{CO2}=0,1.22,4=2,24l\)
c,\(n_{NaCl}=2n.Na2CO3=0,2\left(mol\right)\)
\(\Rightarrow m_{NaCl}=0,2.58,5=11,7g\)
d,\(V_{dd}saupu:0,2l\)
\(\Rightarrow CM_{NaCl}=\frac{0,2}{0,2}=1\left(M\right)\)
\(CM_{HCl_{du}}=\frac{0,2}{0,2}=1\left(M\right)\)
Na2CO3+2HCl---->2NaCl+H2O+CO2
a) n Na2CO3=10,6/106=0,1(mol)
n HCl=0,2.2=0,4(mol)
---->HCl dư dd sau pư làm QT hóa đỏ
b) Theo pthh
n H2=n Na2CO3=0,1(mol)
V H2=0,1.22,4=2,24(l)
c) Theo othh
n NaCl=2n Na2CO3=0,2(mol)
m NaCl=0,2.58,5=11,7(g)
d) n HCl dư=0,2(mol)
CM HCl= 0,2/0,2=1(M)
CM NaCl=0,2/0,2=1(M)