CHEMICAL EQUATION Calcium hydroxide may be used to neutralize (completely react

Question

CHEMICAL EQUATION Calcium hydroxide may be used to neutralize (completely react with) aqueous hydrochloric acid. Calculate the number of mass, g, of hydrochloric acid that would be neutralized by 0.500 mol of solid calcium hydroxide. 1. Ca (OH) 2 (s) + HCl (aq) CaCl2 (aq) + H2O (l) What mass of sodium hydroxide, NaOH, would be required to produce 8.00 g of the antacid milk of magnesia, Mg (OH)2, by the reaction of MgCl2 with NaOH? 2. MgCl2 (aq) + NaOH (aq) Mg (OH) 2 (s) + NaCl (aq) 3. Metallic iron reacts with 02 gas to produce iron (III) oxide Write and balance the equation. a. b. Calculate the number of g of iron needed to produce 5.00g of product. Barium carbonate decomposes upon heating to barium oxide and carbon dioxide; (a) Write and balance the equation (b) Calculate the number of g of carbon dioxide produced by heating 50.0g of barium carbonate. 4. Assume that the theoretical yield of iron in the process was 30.0 g. If the actual yield of iron was 25.0g, calculate the percent yield. 5. 2Al (s) + Fe2O3 (s) Al2O3 (I) + 2 Fe(l)

Explanation / Answer

1. Balanced chemical equation is

Ca(OH)2(s) + 2 HCl(aq) -----> CaCl2(aq) + 2 H2O(l)

1 mole of Ca(OH)2 is neutralized by 2 moles of HCl

0.500 mole of Ca(OH)2 is neutralized by ---? moles of HCl = (0.5/1)x2 = 1 mole

1 mole of HCl = 36.5 grams (mass of 1 mole HCl)

2. Balanced chemical equation is

MgCl2(aq) + 2 NaOHl(aq) -----> Mg(OH)2(s) + 2 NaCl(aq)

To produce 1 mole of Mg(OH)2, 2 moles of NaOH is required

To produce 58.3 grams (1 mole) of Mg(OH)2, 80 grams (2 moles) of NaOH is required

To produce 8 grams of Mg(OH)2,---? grams of NaOH is required = (8/58.3) x 80 = 10.98 grams

3.

a)Balanced chemical equation is

4 Fe + 3 O2 ----> 2 Fe2O3

b) 2 mole of Fe2O3 is produced from 4 moles of Fe

   2x 159.7 grams of Fe2O3 is produced from 4x 55.85 grams of Fe

5 grams of Fe2O3 is produced from ----? grams of Fe = (5/2x159.7) x 4 x 55.85 = 3.5 grams

4.

a)Balanced chemical equation is

BaCO3 ----> BaO + CO2

1 mole of BaCO3 produce 1 mole of CO2

197.3 grams of BaCO3 produce 44 grams of CO2

50 grams of BaCO3 produce ---? grams of CO2 =(50/197.3) x 44 = 11.15 grams

5. Theoretical yield = 30 grams

Actual yield = 25 grams

Percent yield = (25/30) x 100 = 83.3%

   

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