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2 KOH (s) + H 2 SO 4 (g)  K 2 SO 4 (aq) + 2 H 2 O (g)

2 KOH (s) + H 2 SO 4 (g)  K 2 SO 4 (aq) + 2 H 2 O (g) In a particular experiment, the reaction of 100. g KOH with 100.g H 2 SO 4 1) Calculate the MASS of all species present at the end of the reaction. FIRST FIND YOUR LIMITING REAGENT. MOLES = MASS/GFM KOH. MOLES = 100.g / 55.99g/mol.

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2 KOH (s) + H 2 SO 4 (g)  K 2 SO 4 (aq) + 2 H 2 O (g)

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  1. 2KOH (s) + H2SO4 (g)K2SO4 (aq) + 2H2O (g) • In a particular experiment, the reaction of 100. g KOH with 100.g H2SO4 • 1) Calculatethe MASS of all species present at the end of the reaction. • FIRST FIND YOUR LIMITING REAGENT MOLES = MASS/GFM KOH MOLES = 100.g / 55.99g/mol MOLES = 1.786033 mol KOH MOLES = MASS/GFM H2SO4 MOLES = 100. g / 98.00g/mol MOLES = 1.0204mol H2SO4 =1= 1.0204 KOH2 1.786033 .50 < .57 The actual ratio is larger than the theoretical, therefore the actual ratio fraction is too large, the denominator is to small, KOH limiting.

  2. ICE CHART SOLUTION SKILL • THE ICE CHART WILL ALLOW YOU TO CALCULATE MOLES OF ALL SPECIES AFTER THE REACTION IS COMPLETED. • MOLES OF ALL PRODUCTS. • MOLES OF ALL EXCESS REACTANTS REMAINNING. • THE FASTEST WAY TO APPROACH A REACTION IN WHICH YOU ARE CALCULATING MANY SUBSTANCES AT ONCE.

  3. DETERMINING THE VALUE OF “X” • X IS DETERMINED BY THE LIMITING REAGENT. • IN THIS PROBLEM KOH, 1.786033 mol = 2X FROM THE LIMITING REAGENT KOH. X = 0.893 MOL. • Quantities of all species are calculated from x: • THE EXCESS REAGENT IS SULFURIC ACID, WHICH IS 1.0204 –X • 1.0204 – 0.893 • 0.1274 MOLE OF H2SO4 REMAINS • 2. K2SO4 PRODUCED IS X, 0.893 MOLE • 3 WATER IS 2X, 2(0.893) = 1.786 MOLE. MOLES = MASS/GFM K2SO4 0.893mol= MASS/173.99 g MASS = 153.87 g K2SO4 154. g MOLES H2O= MASS/GFM H2O 1.786 MOLE = MASS/17.99 : MASS = 32.103g = 32.1g

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