Consider the reaction represented with the following equation Na2Co3+2Hcl ———> 2Nacl +H2O+CO2. What volume of 0.02 moledm-3 Na2(03Ca0) would be require to completely neutraize 40cm3 of 0.10 moldm-3?
n=c×Vn=c\times Vn=c×V
n(HCl)=0.10×0.040=0.004moln(HCl) = 0.10\times 0.040 = 0.004 moln(HCl)=0.10×0.040=0.004mol
According to equation n(Na2CO3):n(HCl)=1:2n(Na_2CO_3):n(HCl) = 1:2n(Na2CO3):n(HCl)=1:2 , then n(Na2CO3)=n(HCl)2=0.0042=0.002moln(Na_2CO_3) =\frac{n(HCl)}{2} = \frac{0.004}{2}= 0.002 moln(Na2CO3)=2n(HCl)=20.004=0.002mol
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