Need help with questions 2-6 Answer the questions in order in the exam book. You

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Need help with questions 2-6

Answer the questions in order in the exam book. You must return the exam paper. A compound copper and sulfur only was found to contain 79.8% copper, what is the empirical formula of the compound? Balance the equation below then calculate how many grams of carbon dioxide can be obtained by burning 4.00 g of carbon with 4.00 g of oxygen: C(s) O_2(g) rightarrow CO_2(g) H_2O(g) Balance each of the following equations: CacO_3(s) HC(aq) rightarrow CaCI_2(aq) CO_2 + H_2O(f) BaCl_2(aq) + Na_2SO_4(aq) rightarrow BaSO_4(s) + NaCl(aq) Write a net balanced ionic equation for the reactions in Q.3 above. A 0.5035 g of potassium hydrogen phthalate (HC_3H_4O_4, a weak monoprotic acid) required 22.55 mL of a NaOH solution for titration to a phenolphthalein endpoint. What is the molarity the solution? When 25.0 mL of 1.00 M HCI is mixed with exactly 25.0 mL of NaOH in a calorimeter, the temperature of the solution increased from 25.0 degree c to 27.55 degree C. If the specific heat the solution is 4.18 J/(g degree C), what is Delta H for the reaction in kJ/mol. Take the total volume of the solution as 50.0 mL of density of the solution as 1.00 g/mL.

Explanation / Answer

1) 79.8 g Cu (1mol Cu/63.546g) = 1.256 mol Cu

0.1245g S (1mol S/32.065g) = 0.63 mol S

mol Cu/mol S = 1.256 /0.63 = 2.0

The ratio of Cu to S is about 2/1

Empirical formula of copper sulfur compound : Cu2S

2) C(s) + O2 (g) ---> CO2 (g) + H2O (g)

The balanced equation: C (s) + O2 (g) ---> CO2 (g)

Therefore, one mole of C reacts with one mole of O2 and produces one mole of CO2;

no. moles of C = 4/12 = 0.333 moles
no. moles of O2 = 4/32 = 0.125 moles

So, the O2 is the limiting reactant, therefore, no. of moles of CO2 = moles of O2 = 0.125 moles

Wt. of CO2 produced = no. of moles x mol.wt = 0.125 x 44 = 5.5 g

3) balaned reactions;

a) CaCO3(s) + 2 HCl(aq) = CaCl2(aq) + CO2(g) + H2O(l)

b) BaCl2(aq) + Na2SO4(aq)---> BaSO4(s) + 2NaCl(aq)

c) Al(OH)3 + 3HNO3(aq) ---> Al(NO3)3 + 3H2O

4) Net ionic equations;

a) CaCO3(s) + 2 H^+(aq) --> Ca^2+(aq) + H2O(l) + CO2(g)
b) Ba2+(aq) + SO42-(aq)---->BaSO4 (s)
c) 3H+ + Al(OH)3 (s) --> Al3+ + 3H2O (l)

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