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Chapter 4: Problem 53

Write balanced molecular and net ionic equations for the reactions of (a) manganese with dilute sulfuric acid, (b) chromium with hydrobromic acid, (c) tin with hydrochloric acid, (d) aluminum with formic acid, HCOOH.

Short Answer

Expert verified
The short version of the answer is as follows: a) Balanced molecular equation: Mn + H2SO4 -> MnSO4 + H2 Net ionic equation: Mn(s) + 2H+(aq) -> Mn^2+(aq) + H2(g) b) Balanced molecular equation: 2Cr + 6HBr -> 2CrBr3 + 3H2 Net ionic equation: 2Cr(s) + 6H+(aq) -> 2Cr^3+(aq) + 3H2(g) c) Balanced molecular equation: Sn + 2HCl -> SnCl2 + H2 Net ionic equation: Sn(s) + 2H+(aq) -> Sn^2+(aq) + H2(g) d) Balanced molecular equation: 2Al + 6HCOOH -> 2Al(HCOO)3 + 3H2 Net ionic equation: 2Al(s) + 6H+(aq) -> 2Al^3+(aq) + 3H2(g)

Step by step solution

01

Reaction a: Manganese with dilute sulfuric acid

First, identify the products of the reaction: Mn + H2SO4 -> MnSO4 + H2 Now, balance the equation: Mn + H2SO4 -> MnSO4 + H2 (already balanced) The balanced molecular equation for the reaction of manganese with dilute sulfuric acid is: Mn + H2SO4 -> MnSO4 + H2 To find the net ionic equation, split all species into their ions: Mn(s) + 2H+(aq) + SO4^2-(aq) -> Mn^2+(aq) + SO4^2-(aq) + H2(g) Canceling out the spectator ions (SO4^2-), we get: Mn(s) + 2H+(aq) -> Mn^2+(aq) + H2(g) The net ionic equation for the reaction of manganese with dilute sulfuric acid is: Mn(s) + 2H+(aq) -> Mn^2+(aq) + H2(g)
02

Reaction b: Chromium with hydrobromic acid

First, identify the products of the reaction: Cr + HBr -> CrBr3 + H2 Now, balance the equation: 2Cr + 6HBr -> 2CrBr3 + 3H2 The balanced molecular equation for the reaction of chromium with hydrobromic acid is: 2Cr + 6HBr -> 2CrBr3 + 3H2 To find the net ionic equation, split all species into their ions: 2Cr(s) + 6H+(aq) + 6Br^-(aq) -> 2Cr^3+(aq) + 6Br^-(aq) + 3H2(g) Canceling out the spectator ions (Br^-), we get: 2Cr(s) + 6H+(aq) -> 2Cr^3+(aq) + 3H2(g) The net ionic equation for the reaction of chromium with hydrobromic acid is: 2Cr(s) + 6H+(aq) -> 2Cr^3+(aq) + 3H2(g)
03

Reaction c: Tin with hydrochloric acid

First, identify the products of the reaction: Sn + HCl -> SnCl2 + H2 Now, balance the equation: Sn + 2HCl -> SnCl2 + H2 The balanced molecular equation for the reaction of tin with hydrochloric acid is: Sn + 2HCl -> SnCl2 + H2 To find the net ionic equation, split all species into their ions: Sn(s) + 2H+(aq) + 2Cl^-(aq) -> Sn^2+(aq) + 2Cl^-(aq) + H2(g) Canceling out the spectator ions (Cl^-), we get: Sn(s) + 2H+(aq) -> Sn^2+(aq) + H2(g) The net ionic equation for the reaction of tin with hydrochloric acid is: Sn(s) + 2H+(aq) -> Sn^2+(aq) + H2(g)
04

Reaction d: Aluminum with formic acid

First, identify the products of the reaction: Al + HCOOH -> Al(HCOO)3 + H2 Now, balance the equation: 2Al + 6HCOOH -> 2Al(HCOO)3 + 3H2 The balanced molecular equation for the reaction of aluminum with formic acid is: 2Al + 6HCOOH -> 2Al(HCOO)3 + 3H2 To find the net ionic equation, split all species into their ions: 2Al(s) + 6H+(aq) + 6HCOO^-(aq) -> 2Al^3+(aq) + 6HCOO^-(aq) + 3H2(g) Canceling out the spectator ions (HCOO^-), we get: 2Al(s) + 6H+(aq) -> 2Al^3+(aq) + 3H2(g) The net ionic equation for the reaction of aluminum with formic acid is: 2Al(s) + 6H+(aq) -> 2Al^3+(aq) + 3H2(g)

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