Answer to Question #100218 in General Chemistry for David Jeremiah

Question #100218

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?


1
Expert's answer
2019-12-10T07:44:29-0500
Na2CO3+2HCl=2NaCl+H2O+CO2Na_2CO_3+2HCl = 2NaCl + H_2O + CO2

n=c×Vn=c\times V

n(HCl)=0.10×0.040=0.004moln(HCl) = 0.10\times 0.040 = 0.004 mol

According to equation n(Na2CO3):n(HCl)=1:2n(Na_2CO_3):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 mol


V(Na2CO3)=nc=0.0020.02=0.1L=100mlV(Na_2CO_3) = \frac{n}{c} = \frac{0.002}{0.02} = 0.1 L = 100 ml


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment