Answer to Question #100218 in General Chemistry for David Jeremiah

Question

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?

Answer 1

$$Na_2CO_3+2HCl = 2NaCl + H_2O + CO2$$

$n=c\times V$

$n(HCl) = 0.10\times 0.040 = 0.004 mol$

According to equation $n(Na_2CO_3):n(HCl) = 1:2$ , then $n(Na_2CO_3) =\frac{n(HCl)}{2} = \frac{0.004}{2}= 0.002 mol$

$$V(Na_2CO_3) = \frac{n}{c} = \frac{0.002}{0.02} = 0.1 L = 100 ml$$