1)Consider the reaction when aqueous solutions of magnesium nitrate and cobalt(I

ID: 480419 • Letter: 1

Question

1)Consider the reaction when aqueous solutions of magnesium nitrate and cobalt(II) sulfate are combined.

What is the net ionic equation?

2) For the following reaction, 48.8 grams of potassium hydrogen sulfate are allowed to react with 23.3 grams of potassium hydroxide.

potassium hydrogen sulfate (aq) + potassium hydroxide (aq) potassium sulfate (aq) + water (l)

What is the maximum amount of potassium sulfate that can be formed? grams

What is the FORMULA for the limiting reagent?

What amount of the excess reagent remains after the reaction is complete? grams

Mg(NO3)2+ CoSO4---->MgSO4 + Co(NO3)2 (s)molecular reaction

Mg+2 + 2NO3- +CO+2 +SO4-2 ----------->Mg+2 +SO4-2 +CO(NO3)2

CO+2 +2NO3- ---->Co(NO3)2 ( net ionic equation)

2.potassium hydrogen sulfate (aq) + potassium hydroxide (aq) potassium sulfate (aq) + water (l)

KOH +KHSO4---->K2SO4 +H2O, molar masses : KOH= 56, KHSO4=136, K2SO4= 174, H2O= 18

1 mole of KOH reacts with 1 mole of KHSO4 to form 1 mole of K2SO4

Molar ratio of reactants ( as per the stoichiometry)= 1:1

Moles of reactants given : Mass/Molar mass,

Moles : KOH= 23.3/56=0.42, KHSO4= 48.8/136=0.36

molar ratio of reactants given : 0.42:0.36 = 1.2: 1

So excess reactant is KOH and limiting reactants is KHSO4. Moles of K2SO4 formed =0.36 ( this is the maximum)

Mass of KHSO4 formed = 0.36*136=48.96 gm

Excess reactant remaining = 0.42-0.36= 0.06 moles of KOH. mass of KOH remaining =56*0.06=3.36 gm

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