Question

составить 6 уравнений ионного обмена, полное ионное, сокращённое ионное

Answer 1

$begin{array}{l} mathrm{AgNO_{3}+NaCl=AgCl+NaNO_{3}} \ mathrm{Ag^{+}+NO_{3}^{-}+Na^{+}+Cl^{-}=AgCl+Na^{+}+NO_{3}^{-}} \ mathrm{Ag^{+}+Cl^{-}=AgCl} end{array}$.

$begin{array}{l} mathrm{H_{2}SO_{4}+BaCl_{2}=BaSO_{4}+2HCl} \ mathrm{2H^{+}+SO_{4}^{2-}+Ba^{2+}+2Cl^{-}=BaSO_{4}+2H^{+}+2Cl^{-}} \ mathrm{Ba^{2+}+SO_{4}^{2-}=BaSO_{4}} end{array}$.

$begin{array}{l} mathrm{CuSO_{4}+H_{2}S=CuS+H_{2}SO_{4}} \ mathrm{Cu^{2+}+SO_{4}^{2-}+2H^{+}+S^{2-}=CuS+2H^{+}+SO_{4}^{2-}} \ mathrm{Cu^{2+}+S^{2-}=CuS} end{array}$.

$begin{array}{l} mathrm{PbCl_{2}+2KI=PbI_{2}+2KCl} \ mathrm{Pb^{2+}+2Cl^{-}+2K^{+}+2I^{-}=PbI_{2}+2K^{+}+2Cl^{-}} \ mathrm{Pb^{2+}+2I^{-}=PbI_{2}} end{array}$.

$begin{array}{l} mathrm{CaCl_{2}+K_{2}CO_{3}=CaCO_{3}+2KCl} \ mathrm{Ca^{2+}+2Cl^{-}+2K^{+}+CO_{3}^{2-}=CaCO_{3}+2K^{+}+2Cl^{-}} \ mathrm{Ca^{2+}+CO_{3}^{2-}=CaCO_{3}} end{array}$.

$begin{array}{l} mathrm{Na_{2}CO_{3}+2HCl=2NaCl+CO_{2}+H_{2}O} \ mathrm{2Na^{+}+CO_{3}^{2-}+2H^{+}+2Cl^{-}=2Na^{+}+2Cl^{-}+CO_{2}+H_{2}O} \ mathrm{CO_{3}^{2-}+2H^{+}=CO_{2}+H_{2}O} end{array}$.