7 Calcium carbonate decomposes at high temperatures to form

7. Calcium carbonate decomposes at high temperatures to form carbon dioxide and calcium oxide: a) Write the balanced chemical equation for this reaction below b) How many grams of calcium carbonate will I need to form 3.45 liters of carbon dioxide?

Solution

7.The reaction decomposition of CaCO3 is as follows

a) CaCO3(s) -------------------- CaO(s)   + CO2(g)

     1 mole                                             1 mole

1 mole of gas ocupies 22.4 L at STP

molar mass of CaCO3 = 100 gram/mole

according to equation

1 mole of CO2= 1 mole of CaCO3

22,4 L of CO2 = 100 grams of CaCO3

3.45 L of CO2 = ?

                         = 100x3.45/22.4 = 15.4 grams

mass of CaCO3 required = 15.4 grams

9.

Na2S2O3(aq)   + 4 Cl2(g) + 5 H2O(aq) ------------ 2 NaHSO4 (aq) +8 HCl(aq)

a) according to equation

4 moles of Cl2 = 1 mole of Na2S2O3

0.12 mole of Cl2 = ?

                       = 1 x0.12/4 = 0.03 moles of Na2S2O3

number of moles of Na2S2O3 required = 0.03 moles

b)according to equation

4 moles of Cl2 = 8 moles of HCl

0.12 moles of Cl2 = ?

                       = 8x0.12/4 = 0.24 moles of HCl

number of moles of HCl formed = 0.24 moles

c) according to equation

4 moles of Cl2 = 5 moles of H2O

0.12 moles of Cl2 = ?

                              = 5 x 0.12/4 =0.15 moles of H2O

number of moles of H2O required = 0.15 moles

d) According to equation

8 moles of HCl= 5 moles of H2O

0.24 moles of HCl= ?

                             = 5 x 0.24/8 = 0.15 moles of H2O

number of moles of H2O required = 0.15 moles.