Answer the questions in order in the exam book·H, L0; O 16.0; N, 14.0; Zn 65.3,

Question

Answer the questions in order in the exam book·H, L0; O 16.0; N, 14.0; Zn 65.3, S-32.0; C, 35.5 1. A compound of zinc and chlorine only was found to contain 47.9 % zinc. Use the data toenpirical formula of the compound and write a balanced equation showing how the compound is formed 2. Balance the equation below then calculate how many grams of carbon dioxide can be obtained by burning 6.00 g of C2Hs in 6.00 g of oxygen: 3. Complete and balance each of the following equations: a) Nazcols) + HCl(aq) c) Al(OH)1(s) + H2SOdoq) 4. Write a net balanced ionic equation for each of the reactions in Q3 above. 5. Write a balanced equation for the reaction of sulfamic acid (H2NSO3H) and sodium hydroxide then calculate the molarity of a NaOH solution if 0.5026 g of the acid required 20.55 mL of a NaOH solution for titration to a phenolphthalein endpoint. 6. Use the data below to calculate the enthalpy of reaction for: NaOH(s)+H'(aq)+Cr(aq) HOU) +Na(aq)+ Cr(aq) (1) A:-33.44k/mol NaOH(s) Na.(aq) + OH(aq) H'(aq) + crtaq) + Na"(aq) + OH"(aq) NaTaq) + cr(aq) + Hon. (2) A:-33.44k/mol Is the reaction exothermic or endothermic? Briefly explain your answer.

Explanation / Answer

1)

We can solve this by using this table.First find mol by dividing % by atomic mass.The take the least value from the mol and find the mole ratio using this value.

Element % mol = %/atomic mass mol ratio

Zn 47.9 47.9/65.3 = 0.73 0.73/0.73 = 1

Cl 52.1 52.1/35.5= 1.46 1.46/0.73 = 2

Hence mole ratio of Zn : Cl is 1 : 2. So empirical formula = ZnCl2.

Balanced standard equation for formation of ZnCl2 is,

Zn (s) + Cl2 (g) --------> ZnCl2(s).

But ZnCl2 is also formed by reaction of Zn on HCl.

Zn(s) + 2 HCl(aq) --------> ZnCl2 (aq) + H2 (g).

2)

2 C2H6 (g) + 7 O2 (g) ------> 4 CO2 (g) + 6 H2O(l)

Thus according to stoichiometric equation 2 moles C2H6 on combustion with 7 moles of O2 to yields 4 moles CO2.

C2H6 Molecular weight = (2*12)+ (6*1) = 30.0 g

Mol of C2H6 = 6.00/30.0 = 0.2

O2 molecular mass = (2*16) = 32

Mol of O2 = 6.00/32 = 0.19

Thus 0.2 moles C2H6 need 0.7 moles of O2 for complete reaction. So here O2 is limiting reagent. Limiting reagent means reagent which is completely consumed during reaction.

Thus 0.19 moles O2 requires =(2*0.19)/7 = 0.05 mol C2H6.

2 mol C2H6 produces 4 mol CO2.

Hence 0.05 mol C2H6 produces 0.10 mol CO2 .

1 mol CO2 = 44.0 g

So 0.1 mol CO2 = 4.4 g

Thus 6.0 g C2H6 and 6.0 g O2 produces 4.4 g CO2.

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