A) Write the names and chemical formula for the strong acids and bases. B) A 30.

Question

A) Write the names and chemical formula for the strong acids and bases.

B) A 30.4 g sample of an oxygen-containing hydrocarbon (CxHyOz) is combusted and yields 70.4 g CO2 and 14.4 g H2O. Its molecular mass is 152 g/mol. a. Determine both the empirical and molecular formulas. b. Determine the mass (g) of O2 used in the reaction.

C) A 44.23 g sample of metallic copper is added to 136 mL of 6.0 M nitric acid to produce aqueous copper(II) nitrate, liquid water and gaseous nitrogen monoxide. a. Write a balanced chemical equation. b. What is the limiting reagent? c. How many grams of nitrogen monoxide are produced?

D) a. What mass of BaBr2 must be added to 1.00*102 mL of 0.105 M HBr so that the resulting solution has a molarity of 0.150 M in Br? Assume the volume remains the same. b. If 2.00*102 mL of water is added to the 0.150 M solution of bromide ion, what is the final molarity of bromide ion in the solution?

Explanation / Answer

a)

Strong acid means which readily give H+ ion and strong base means which can give OH- ions readily.

Examples for acids;

Sulphuric acid - H2SO4

PerChloric acid - HClO4

Hydrochloric acid - HCl

Examples for bases;

Sodium hydroxide - NaOH

Potassium Hydroxide - KOH

Magnesium hydroxide - Mg(OH)2

B)

CxHyOz + O2 ------> x CO2 + y/2 H2O

from the above stochiometric equation,

1 mole of CO2 requires 1 mole of the compound

44g of CO2 requires 12g of C in the compound

70.4 g of CO2 requires 12/44*70.4 g of C in the compound

                      = 19.2 g of the C

Percentage of C = mass of the c in the compound/mass of the compound *100 = 19.2/30.4*100 = 63.16%

1 mole of H2O requires 2 mole of the compound

18g of H2O requires 2g of H in the compound

14.4 g of H2O requires 2/18*14.4 g of H in the compound

                      = 1.6 g of the H

Percentage of H = mass of the H in the compound/mass of the compound *100 = 1.6/30.4*100 = 5.26%

Percentage of Oxygen = 100 - (percentage of all elements)

Percentage of Oxygen = 100 - (63.16 + 5.26) = 31.58%

Hence the empirical formula = C8H8O3

Empirical mass = 8*12 + 8 *1 + 8*16 = 152 g/mol

number of empirical units = molecular mass/empirical mass = 152/152 = 1

molecular formula = C8H8O3

C)

3 Cu + 8 HNO3 (aq) -----> 3 Cu(NO3)2 (aq) + 2 NO + 4 H2O

number of moles of HNO3 = 136 * 10^-3 * 6 = 0.816 moles

number of moles of Cu = mass/molar mass = 44.23/63.55 = 0.696 moles

3 moles of Cu require 8 moles of HNO3

0.696 moles of Cu requires 0.696*8/3 = 1.856 moles of HNO3

so limiting reagent is HNO3

8 moles of HNO3 produces 2 moles of NO

0.816 moles of HNO3 produces 2/8*0.816 moles of NO

                             = 0.204 moles of NO

mass of NO produced = 0.204 * 30 g/mole = 6.12 g of NO produced.

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