7.
\(n_{CO_2}=\dfrac{11.2}{22.4}=0.5\left(mol\right)\)
\(n_{NaOH}=\dfrac{500\cdot1.3\cdot2.5\%}{40}=0.40652\left(mol\right)\)
\(T=\dfrac{0.40625}{0.5}=0.8125\)
=> Tạo muối axit
\(NaOH+CO_2\rightarrow NaHCO_3\)
\(0.40625.......................0.40625\)
\(C_{M_{NaHCO_3}}=\dfrac{0.40652}{0.5}=0.8125\left(M\right)\)
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8.
\(n_{CO_2}=\dfrac{0.448}{22.4}=0.02\left(mol\right)\)
\(n_{NaOH}=0.1\cdot0.25=0.025\left(mol\right)\)
\(T=\dfrac{0.025}{0.02}=1.25\)
=> Tạo ra 2 muối
\(n_{Na_2CO_3}=a\left(mol\right),n_{NaHCO_3}=b\left(mol\right)\)
Ta có :
\(2a+b=0.025\)
\(a+b=0.02\)
\(\Rightarrow a=0.005,b=0.015\)
\(m_{Muối}=0.005\cdot106+0.015\cdot84=1.79\left(g\right)\)
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