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Lecture 7sf

# Lecture 7sf - Stoichiometry of Precipitation Reactions What...

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Stoichiometry of Precipitation Reactions What mass of NaCl is needed to react completely with 1.50 L of 0.100 M AgNO 3 ? AgNO 3 + NaCl → AgCl + NaNO 3 Net ionic: Ag + (aq) + Cl - (aq) → AgCl(s) 1.50 L x 0.100 x 3 AgNO mol 1 NaCl mol 1 x 58.45 = 8.77 g NaCl Mix: 1.25 L of 0.0500 M Pb(NO 3 ) 2 2.00 L of 0.0250 M Na 2 SO 4 What mass of PbSO 4 forms? What amount of what reactant is left over? Pb(NO 3 ) 2 + Na 2 SO 4 → PbSO 4 + 2NaNO 3 Net ionic: Pb 2+ (aq) + SO 4 2- (aq) PbSO 4 (s) 1.25 L x 0.0500 = 0.0625 mol Pb(NO 3 ) 2 2.00 L x 0.0250 = 0.0500 mol Na 2 SO 4 Since the stoichiometric ratio is 1:1, the lower number, 0.0500 mol Na 2 SO 4 is the limiting reactant. That is 0.0500 mol Na 2 SO 4 reacts with 0.0500 mol Pb(NO 3 ) 2 . Excess: (0.0625 - 0.0500) mol Pb(NO 3 ) 2 0.0125 mol of Pb(NO 3 ) 2 is left over. To find the mass PbSO 4 formed, work with the limiting reactant. 0.0500 mol Na 2 SO 4 x 4 2 4 SO Na mol 1 PbSO mol 1 x 303 = 15.2 g PbSO 4 formed Stoichiometry of Acid-Base Reactions

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What volume of 0.250 M HCl is required to react completely with 20.0 mL of 0.500 M NaOH? HCl + NaOH → NaCl + HOH Net ionic: H + (aq) + OH - (aq) → HOH 0.020 L x 0.500 x NaOH mol 1 HCl mol 1 x = 0.040 L or 40.0 mL HCl What volume of 0.250 M H 2 SO 4 is required to react completely with 20.0 mL of 0.500 M NaOH? H 2 SO 4 + 2NaOH → Na 2 SO 4 + 2HOH Net ionic: H + (aq) + OH - (aq) → HOH 0.020 L x 0.500 x NaOH mol 2 SO H mol 1 4 2 x = 0.020 L or 20.0 mL H 2 SO 4 Compare the previous two examples, which were identical except for using 0.250 M H 2 SO 4 instead of 0.250 M HCl. Since H 2 SO 4 has two ionizable hydrogens, there is a 1:2 ratio of H 2 SO 4 with NaOH, meaning that 20 mL of 0.250 M H 2 SO 4 will react with the same amount of NaOH as 40 mL of 0.250 M HCl. What amount of water is formed when 25.0 mL of 0.200 M HNO 3 is mixed with 40.0 of 0.100 M NaOH? HNO 3 + NaOH → NaNO 3 + HOH Net ionic: H + (aq) + OH - (aq) HOH 0.0250 L x 0.500 = 0.00500 mol HNO 3 0.0400 L x 0.100 = 0.00400 mol NaOH Since stoichiometric ratio between HNO 3 and NaOH is 1:1, the NaOH is the limiting reactant. Since the stoichiometric ratio between NaOH and H 2 O is also 1:1, 0.00400 mol H 2 O is formed. What mass of Ca(OH) 2 is needed to neutralize 275 mL of 0.600 M HCl? Ca(OH) 2 + 2HCl → CaCl 2 + 2HOH
0.275 L x x HCl mol 2 ) OH ( Ca mol 1 2 x 74.1 = 6.11 g Ca(OH) 2 A 1.3009 g mass of potassium hydrogen phthalate, KHC 8 H 4 O 4 (molar mass 204.22) neutralizes 41.20 mL of NaOH solution. What is the molarity of the NaOH? Assume that the potassium hydrogen phthalate has one ionizable hydrogen (reacts with NaOH in a 1:1 mole ratio).

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Lecture 7sf - Stoichiometry of Precipitation Reactions What...

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