Bài 1 :
$a) SO_3 + H_2O \to H_2SO_4 \\ b) CO_2 + H_2O \rightleftharpoons H_2CO_3 \\ c) P_2O_5 + 3H_2O \to 2H_3PO_4 \\ d) CaO + H_2O \to Ca(OH)_2 \\ e) Na_2O + H_2O \to 2NaOH$
Bài 2 :
\(n_{H_2SO_4}=0.3\cdot1.5=0.45\left(mol\right)\)
\(a.\)
\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O\)
\(0.9..............0.45\)
\(m_{NaOH}=0.9\cdot40=36\left(g\right)\)
\(\Rightarrow m_{dd_{NaOH}}=\dfrac{36}{40\%}=90\left(g\right)\)
\(b.\)
\(n_{NaOH}=n_{KOH}=0.9\left(mol\right)\)
\(m_{KOH}=0.9\cdot56=50.4\left(g\right)\)
\(\Rightarrow m_{dd_{KOH}}=\dfrac{50.4}{5.6\%}=900\left(g\right)\)
\(\Rightarrow V_{dd_{KOH}}=\dfrac{900}{1.045}=861.2\left(ml\right)\)
Bài 1 :
\(SO_3+H_2O\rightarrow H_2SO_4\)
\(CO_2+H_2O⇌H_2CO_3\)
\(P_2O_5+3H_2O\rightarrow2H_3PO_4\)
\(CaO+H_2O\rightarrow Ca\left(OH\right)_2\)
\(Na_2O+H_2O\rightarrow2NaOH\)
Bài 2 :
$a) 2NaOH + H_2SO_4 \to Na_2SO_4 + H_2O$
$n_{H_2SO_4} = 0,3.1,5 = 0,45(mol)$
$n_{NaOH} = 2n_{H_2SO_4} = 0,9(mol)$
$m_{dd\ NaOH} = \dfrac{0,9.40}{40\%} = 90(gam)$
$b) n_{KOH} = n_{NaOH} = 0,9(mol)$
$m_{dd\ KOH} = \dfrac{0,9.56}{5,6\%} = 900(gam)$
$V_{dd\ KOH} = \dfrac{900}{1,045} = 861,24(ml)$
Bài 3 :
Đặt : CTHH của muối cacbonat có dạng : \(MCO_3\)
\(MCO_3+2HCl\rightarrow MCl_2+CO_2+H_2O\)
\(M+60................M+71\)
\(12.4......................16\)
Khi đó :
\(16\cdot\left(M+60\right)=12.4\cdot\left(M+71\right)\)
\(\Rightarrow M=-22.11\)
=> Sai đề