s10q08 - 1 Chem 171 Quiz 8 Version A 12.0 g of NaOH is...

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Chem 171 Quiz 8 Version A 1. 12.0 g of NaOH is required to neutralize 400. cm 3 of a solution of HCl in water. Na = 22.99. O = 16.00. H = 1.008 Write the molecular equation. Calculate the concentration of the HCl solution in M. Calculate the concentration of HCl solution in N. NaOH (aq) + HCl (aq) NaCl (aq) + H 2 O (l) Moles of NaOH = (12.0 g NaOH)(1 mol)(40.00g) = 0.300 mol NaOH 0.300 moles of NaOH = 0.300 moles of HCl 0.300 moles of HCl / 0.400 L = 0.75 M 0.75 M = 0.75 N 2. The following questions pertain to the experiment entitled” Evaluating Commercial Antacids”. Write a molecular equation for the reaction of magnesium hydroxide with hydrochloric acid. Show the states. Write the molecular equation for the neutralization of calcium carbonate antacid. Show states. Mg(OH) 2 (s) + 2 HCl(aq) MgCl 2 (aq) + 2H 2 O (l) CaCO 3 (s) + 2 HCl(aq) CaCl 2 (aq) + CO 2 (g) + H 2 O(l) or H 2 CO 3 (aq) Chem 171 Quiz 8 Version B 1. What is the definition of neutralization? What mass of KOH will be required for the preparation of 500. cm 3 of 0.400 N KOH for use in neutralization reactions? K = 39.10. O = 16.00. H = 1.008. A neutralization is a reaction between an acid and a base. 0.400 N KOH = 0.400 M KOH 0.400 M KOH = 0.400 moles of KOH / liter of solution 0.500 L x 0.400 mole of KOH / liter = 0.200 mol KOH 0.200 mol KOH x 56.1 g / mol = 11.2 g KOH 2.
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