Question

1) Write the balanced neutralization reaction between H2SO4 and KOH in an aqueous solution.

0.350L of 0.430 M H2SO4 is mixed with 0.300 L of 0.270 M KOH what concentration of sulfuric acid remans after neutralization?

2) Balance the following: AgNO3 (aq) + NaCL (aq) ----->

                                         K3PO4 (aq) + MgCL2 (aq) ----->

3) Complete and balance the precipiation reaction: CuCl2 (aq) + Na2CO3(aq) ---->

Answer 1

(1) balaced equation

   H2SO4 (aq)+ 2 KOH (aq) -------------------------> K2SO4 (aq) + 2 H2O (l)

solution :

moles of H2SO4 = 0.350 x .430 = 0.1505

moles of KOH = 0.3 x 0.270 = 0.081

H2SO4 (aq) + 2 KOH (aq) -------------------------> K2SO4 (aq) + 2 H2O (l)

1   mol                   2 mol

0.1505 mol             0.081

for 0.081 mol KOH howmany moles H2SO4 need ?

2 mol KOH ----------------------> 1 mol H2SO4

0.081 mol KOH ------------------> x mol H2SO4

x = 0.081 / 2 = 0.0405 mol H2SO4 needed

we have 0.1505 mol H2SO4 .

remining H2SO4 moles = 0.1505 - 0.0405 = 0.11

H2SO4 moles remaining = 0.11 mol

total volume = 0.35 + 0.3 L = 0.65 L

concentration of H2SO4 = moles / volume

                                        = 0.11 / 0.65

                                       = 0.169 M

after neutralization   concentration of H2SO4 = 0.169 M

2)

AgNO3 (aq) + NaCl (aq) ----------> AgCl (s) + NaNO3 (aq)

2K3PO4 (aq) + 3 MgCl2 (aq) ------------------> Mg3 (PO4)2 (s) + 6 KCl (aq)

3)

CuCl2 (aq) + Na2CO3(aq) -----------------------> CuCO3(s) + 2NaCl(aq)