Question

1. (2 pts) From the balanced molecular equations, write the complete ionic and net ionic equations for the following: (a) K2C204(aq)+Ba(OH)2(aq)-2KOH(aq)+BaC204(s) 2. (4 pts) Determine the oxidation states of the elements in the compounds listed. None of the oxygen-containing compounds are peroxides or superoxides. (a) LiNO3 (b) Mg Si (c) H3PO4 (d) P406 3. (2 pts) Write the balanced equation, then determine: The number of moles and the mass of magnesium carbonate, MgCO3, required to produce 283 g of carbon dioxide. (MgO is the other product.) 4. (2 pts) Determine the limiting reactant when 0.50 mol of Cr and 0.75 mol of H3PO4 react according to the following chemical equation. 2Cr+2H3PO4-2CrPO4+3H2 How many grams of CrPO4 can be made?

Answer 1

1)

a ) complete ionic : 2K+(aq) + C2O42-(aq) + Ba2+(aq) + 2OH-(aq) ------------> 2K+(aq)+ 2OH-(aq) + BaC2O4(s)

b) net ionic equation : C2O42-(aq) + Ba2+(aq) ------------> BaC2O4(s)

2)

a) Li = +1 , N = +5 , O = -2

b)   Mg = +2 , Si = +4

c)   H = +1 , P   = +5 , O = -2

d) P = +3 , O = -2

3)

MgCO3 ------------------------> MgO    +     CO2

84.3 g                                     40.3 g        44 g

?                                                               283 g

mass of MgCO3   = 84.3 x 283 / 44

                             = 542 g

mass of MgCO3 = 542 g

moles of MgCO3 = mass / molar mass

                               = 542 / 84.3

                              = 6.43

moles of MgCO3 = 6.43

4)

moles of CrPO4 formed = 0.50

mass of CrPO4   = moles x molar mass

                           = 0.50 x 147

                           = 73.5 g