下列离子方程式书写正确的是$(\quad\ \ \ \ )$

向$\\text{NaHC}{{\\text{O}}_{3}}$溶液中加入过量的$\\text{Ba}{{\\left( \\text{OH} \\right)}_{2}}$溶液：$2\\text{HCO}_{3}^{-}+\\text{B}{{\\text{a}}^{2+}}+2\\text{O}{{\\text{H}}^{-}}=\\text{BaC}{{\\text{O}}_{3}}\\downarrow +2{{\\text{H}}_{2}}\\text{O}+\\text{CO}_{3}^{2-}$

将二氧化碳通入氯化钙溶液中：$\\text{C}{{\\text{O}}_{2}}+{{\\text{H}}_{2}}\\text{O}+\\text{C}{{\\text{a}}^{2+}}=\\text{CaC}{{\\text{O}}_{3}}\\downarrow +2{{\\text{H}}^{+}}$

稀盐酸去除铁锈：$\\text{F}{{\\text{e}}_{2}}{{\\text{O}}_{3}}+6{{\\text{H}}^{+}}=2\\text{F}{{\\text{e}}^{3+}}+3{{\\text{H}}_{2}}\\text{O}$

${{\\text{H}}_{2}}^{18}\\text{O}$中投入$\\text{N}{{\\text{a}}_{2}}{{\\text{O}}_{2}}$固体：$2{{\\text{H}}_{2}}^{18}\\text{O}+2\\text{N}{{\\text{a}}_{2}}{{\\text{O}}_{2}}=4\\text{O}{{\\text{H}}^{-}}+4\\text{N}{{\\text{a}}^{+}}+{{}^{18}}{{\\text{O}}_{2}}\\uparrow $

$\rm A$．向$\rm NaHCO_{3}$溶液中加入过量的$\rm Ba(OH)_{2}$溶液，不足量的$\rm NaHCO_{3}$完全反应，产生$\rm BaCO_{3}$沉淀、$\rm NaOH$、$\rm H_{2}O$，离子方程式为：$\text{HCO}_{3}^{-}+\text{B}{{\text{a}}^{2+}}+\text{O}{{\text{H}}^{-}}=\text{BaC}{{\text{O}}_{3}}\downarrow +{{\text{H}}_{2}}\text{O}$，$\rm A$错误；

$\rm B$．$\rm HCl$酸性大于$\rm H_{2}CO_{3}$，二氧化碳与氯化钙不反应，$\rm B$错误；

$\rm C$．铁锈的主要成分为$\rm Fe_{2}O_{3}$，稀盐酸去除铁锈：$\text{F}{{\text{e}}_{2}}{{\text{O}}_{3}}+6{{\text{H}}^{+}}=2\text{F}{{\text{e}}^{3+}}+3{{\text{H}}_{2}}\text{O}$，$\rm C$正确；

$\rm D$．${{\text{H}}_{2}}^{18}\text{O}$中投入$\rm Na_{2}O_{2}$固体，二者发生反应：$\rm Na_{2}O_{2}+2{{\text{H}}_{2}}^{18}\text{O}\rm =H_{2}O_{2}+2\text{N}{{\text{a}}^{\text{18}}}\text{O}\rm H$，$\rm 2H_{2}O_{2}=O_{2}\uparrow+2H_{2}O$，因此该反应的离子方程式为：$\rm 4{{\text{H}}_{2}}^{18}\text{O}\rm +2Na_{2}O_{2}=4Na^{+}+2H_{2}O+4{}_{{}}^{18}\text{O}{{\text{H}}^{-}}\rm +O_{2}↑$，$\rm D$错误；

故选：$\rm C$