Question

How many grams of calcium phosphate (MM=310) will be formed if 25.0 mL of 0.095 M Ca(OH)2 is added to 20.0 mL of 0.104 M phosphoric acid? 3 Ca(OH)2 + 2 H3PO4 

7. (1.5)Complete the following chemical reaction and give which element is oxidized. Cu(s) + 2 AgCl(aq)  + the element or species oxidized is

8. (2) Calculate and compare the oxidation numbers of the Cr in K2CrO4 and in K2Cr2O7.

Answer 1

Number of moles of Ca(OH)2 = 0.095 * 25 * 10^-3 = 0.002375 moles

Number of moles of H3PO4 = 0.104 * 20 * 10^-3 = 0.00208 moles

3 Ca(OH)2 + 2 H3PO4 -------> Ca3(PO4)2 + 6 H2O

Among the both the compounds H3PO4 is limiting reagent since it is present less in number.

2 moles of H3PO4 produces 1 mole of Ca3(PO4)2

0.00208 moles of H3PO4 produces 1/2*0.00208 moles of Ca3(PO4)2

                               = 0.00104 moles

mass of Ca3(PO4)2 = 0.00104 * 310 g/mol = 0.3224 g

7)

The reaction is not possible between Cu and AgCl because the standard reduction potential of Cu is +0.34 and AgCl is +0.22 so the Cu acts as strong Oxidising agent compare to AgCl or AgCl acts as strong reducing agent so the reaction is not possible between Cu and AgCl.

Cu (s) + 2 AgCl (aq) --------> CuCl2 (aq) + 2 Ag (s) (Not possible)

8)

K2Cr2O4

Oxidation state of Cr = 2(+1) + 2(x) + 4(-2) = 0

2+2x-8 = 0

2x = 8-2

x = 3

Oxidation state of Cr = +3 in K2Cr2O4

K2Cr2O7

Oxidation state of Cr = 2(+1) + 2(x) + 7(-2) = 0

2+2x-14 = 0

2x = 14 - 2

x = 6

Oxidation state of Cr = +6 in K2Cr2O7