Astudent synthesized a nickel (ii) ammonia complex with a molecular formula of [ni(nh3)x(h2o)y]clz. the amount of ammonia in the complex was analyzed using 0.2005m hcl. the student determined that the reaction required 20.02ml of hcl to react with 0.1550g of the nickel(ii) ammonia complex. answer the following questions. atomic masses: ni = 58.69g/mol; h= 1.00g/mol; o= 16.00g/mol; cl = 35.45g/mol; n= 14.00g/mol(a) determine the value for "z" in the compound. briefly explain your reasoning. hint: the complex salt must be electrically neutral (i. e. no net charge)(b) calculate the experimental equivalent weigth for the nickel(ii) ammonia complex (i. e. grams of complex per mole of ) based on the results from (a) and (b), propose a molecular formula for the neutral nickel(ii) ammonia coordination complex. show all your work. hint: x+y≤6. try different combinations of "x" and "y" to see which combination best matches the calculated experimental equivalent weight.

Answers explained below

Explanation:

(a)Given,

Molecular formula of the complex = [Ni(NH3)x(H2O)y]Clz

(i) Ni is in +2 oxidation state in the complex.

(ii) NH3 and H2O are the neutral ligands but Cl  is the negatively charged ligand.

(iii) complex is neutral

So, to make the nickel complex in +2 oxidation with neutral charge, we requires 2 Cl-.

Hence, form the above statements, we can say that here in the complex

z=2

(b) Molarity of HCl = 0.2005M

Volume of HCl used = 20.02mL = 20.02*10-3 L

Weight of the nickel(II) ammonia complex = 0.1550g

Reaction of HCl with Ammonia,

HCl (aq) + NH3 (aq) -> NH4Cl (aq)

HCl reacts with ammonia in 1:1 ratio to form ammonium salt (NH4Cl). That means 1 mol of HCl reacts with 1 mol of NH3.

So, we have to find number of moles of HCl used.

No. of moles of HCl used = Molarity of HCl * Volume of HCl used (L)

= 0.2005M * 20.02*10-3 L = 4.014*10-3 moles

Hence no. of moles of ammonia in the complex = No. of moles of HCl used = 4.014*10-3 mol

So, Experimental Equivalent Weight = Weight of the nickel(II) ammonia complex/ No. of moles of NH3

= 0.1550g / 4.014*10-3 mol

= 38.615 g/mol

Hence, Experimental Equivalent Weight = 38.615 g/mol

(c) Given,

x+y=< 6

Molar mass of [Ni(NH3)x(H2O)y]Clz = 58.69 + x(14.00+3*1) + y(16+2*1)+z(35.45)

= (58.69 + 17x + 18y + 35.45z) g/mol

Case1 x=6, y=0 and z=2

Molar mass of [Ni(NH3)6(H2O)0]Cl2   = (58.69 + 17*6 + 18*0 + 35.45*2) g/mol

= 58.69+102+70.90 = 231.59 g/mol

Experimental Equivalent Weight = 233.59/6 = 38.598 g/mol

So, This experimental equivalent weight is equal to the calculated experimental equivalent weight.

Hence the molecular formula of the complex is [Ni(NH3)6]Cl2 where x=6, y=0 and z=2.

Note: You can try other combination but in every case you will find lower or higher calculated experimental equivalent weight.

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the answer is na (sodium)