Question

reactions involving acids
reactants appearance of
reactants observations after mixing
(select all that apply) products
(select all that apply)
na₂co₃(aq) + hcl(aq) the chemicals mixed
together created a
boiling mixer. no observable change
bubbles formed in mixture
mixtures color changed
solid formed in mixture
test tube felt colder
test tube felt hotter no reaction occurred
co₂(g)
h₂co₃(aq)
h₂co₃(g)
h₂o(aq)
h₂o(l)
nacl(aq)
na₂cl(aq)
naoh(aq) + hcl(aq) the test tube chemicals
mix together making it
warm. no observable change
bubbles formed in mixture
mixtures color changed
solid formed in mixture
test tube felt colder
test tube felt hotter no reaction occurred
hocl(aq)
h₂o(aq)
h₂o(l)
nacl(aq)
na₂cl(aq)
nah(aq)
ki(aq) + hno₃(aq) the chemicals
combined together turn
hot. no observable change
bubbles formed in mixture
mixtures color changed
solid formed in mixture
test tube felt colder
test tube felt hotter no reaction occurred
hi(aq)
h₂o(aq)
h₂o(l)
kh(aq)
kno₃(aq)
k₃no(aq)
displacement reactions involving metals

Explanation:

Step1: Analyze the reaction \(Na_{2}CO_{3}(aq)+HCl(aq)\)

The reaction between sodium carbonate (\(Na_{2}CO_{3}\)) and hydrochloric acid (\(HCl\)) is \(Na_{2}CO_{3}+2HCl = 2NaCl + H_{2}O+CO_{2}\uparrow\). Carbon dioxide (\(CO_{2}\)) gas is produced, so bubbles form. The products are \(NaCl(aq)\), \(H_{2}O(l)\) and \(CO_{2}(g)\) (but \(H_{2}CO_{3}\) decomposes to \(H_{2}O\) and \(CO_{2}\) quickly, so we don't consider \(H_{2}CO_{3}(aq)\) or \(H_{2}CO_{3}(g)\) as stable products here).

Step2: Analyze the reaction \(NaOH(aq)+HCl(aq)\)

The reaction is \(NaOH + HCl=NaCl + H_{2}O\). It is a neutralization reaction which is exothermic (test - tube feels hotter). The products are \(NaCl(aq)\) and \(H_{2}O(l)\)

Step3: Analyze the reaction \(KI(aq)+HNO_{3}(aq)\)

There is no chemical reaction between \(KI\) and \(HNO_{3}\) in the sense of a simple double - displacement reaction (since \(KNO_{3}\) and \(HI\) are both soluble). But if we consider redox reactions (assuming no redox conditions are given in the basic experiment description), we assume no reaction. However, if we consider the fact that in some cases (not in basic double - displacement) \(8KI + 5HNO_{3}=8KNO_{3}+4I_{2}+NH_{3}+3H_{2}O\) (but this is a more complex redox reaction). If we assume basic double - displacement (where products should be \(KH\) (not possible as \(H^{-}\) and \(K^{+}\) won't form in aqueous acid) and \(KNO_{3}\), \(HI\) (all soluble)), no reaction in terms of double - displacement (precipitate, gas or weak electrolyte formation). But if we consider the observation of solid formed (assuming some error or different reaction conditions), but in standard double - displacement \(KI + HNO_{3}\to KH + KNO_{3}\) (invalid as \(KH\) is not formed in aqueous acid). If we consider no reaction (in terms of double - displacement), but if we consider the fact that \(HNO_{3}\) is a strong oxidizing agent and \(I^{-}\) is a reducing agent, \(6I^{-}+8H^{+}+2NO_{3}^{-}=3I_{2}+2NO + 4H_{2}O\) (but this is a redox reaction. If we assume the table is for simple acid - base or double - displacement, we assume no reaction (but the solid formed might be due to some other factors). But if we follow the product list, \(KNO_{3}(aq)\) is a possible product (from \(K^{+}\) and \(NO_{3}^{-}\)) and \(HI(aq)\) (from \(H^{+}\) and \(I^{-}\)), but no reaction in terms of double - displacement (since no precipitate, gas or weak electrolyte ( \(HI\) is a strong acid) is formed in a simple double - displacement view. But if we consider the observation of solid formed (maybe due to concentration or other factors not in basic chemistry), but in standard basic chemistry for double - displacement \(KI(aq)+HNO_{3}(aq)\) has no reaction (products \(KNO_{3}(aq)\) and \(HI(aq)\) (both soluble))

Answer:

• For \(Na_{2}CO_{3}(aq)+HCl(aq)\): Observations - Bubbles formed in mixture; Products - \(CO_{2}(g)\), \(H_{2}O(l)\), \(NaCl(aq)\)
• For \(NaOH(aq)+HCl(aq)\): Observations - Test tube felt hotter; Products - \(H_{2}O(l)\), \(NaCl(aq)\)
• For \(KI(aq)+HNO_{3}(aq)\): If we assume no double - displacement reaction (products \(KNO_{3}(aq)\), \(HI(aq)\) (both soluble)), but if we consider the solid formed (assuming some error or non - standard conditions), but in standard basic double - displacement, no reaction (but the problem may expect \(KNO_{3}(aq)\) as a product (from ion combination) and \(HI(aq)\) (but no reaction in terms of double - displacement criteria (precipitate, gas, weak electrolyte)). If we follow the product list and assume some ion - combination (even if no reaction in terms of reaction criteria), \(KNO_{3}(aq)\) is a product. Observations - Solid formed in mixture (assuming non - standard); Products - \(KNO_{3}(aq)\) (if we consider ion combination) and \(HI(aq)\) (but no reaction in terms of double - displacement reaction (since no precipitate, gas or weak acid ( \(HI\) is strong) formation in double - displacement sense)