Write the balanced equation, then outline the steps necessary determine the information equation in each of the following:
(a)the number of moles and the mass of mg required to react with 5.00g of hcl and produce mgcl2 and h2.
(b)the number of moles and the mass of oxygen formed b decomposition of 1.252 g of silver(i) oxide.
(c)the number of moles and the mass of magnesium carbonate, mgco3, required to produce 283 g of carbon dioxide.(mgo is the other product.)
(d) the number of moles and the mass of water formed b combustion of 20.0 kg of acetylene, c2h2, in an excess of oxygen.
(e) the number of moles and mass of berium peroxide, bao2 neede to produce 2.500 kg of barium oxide, bao (o2 is the other product)

a) 3.33 g or 0.137 mol of Mg

b) 0.1729 g or 5.403 mmol (mmol = 1/1000 mol) of silver oxide

c) 542 g or 6.43 mol of magnesium carbonate

d) 13.8 Kg or 768 mol of water

e) 2.761 g or 16.30 mol of barium oxide

Explanation:

a)                                       Mg + 2HCl → MgCl2 + H2

Molar mass             Mg = 24.305 g/mol        HCl = 36.46094 g/molProportions              X g of Mg 5.00 g of HCl

                                         X = 3.33 g of  Mg

Number of moles     3.33 g / 24.305 g/mol = 0.137 mol of Mg

b)                                       2AgO2 → 2Ag + O2

Molar mass             AgO2 = 231.735 g/mol             O2 = 31.998Proportions              1.252 g of AgO2 X g of O2

                                         X = 0.1729 g of  O2

Number of moles    0.1729 g / 31.998 g/mol = 0.005403 mol of O2

c)                                       MgCO3 → MgO + CO2

Molar mass             MgCO3  = 84.3139 g/mol         CO2 = 44.01 g/molProportions              X g ofMgO3 X g of O2

                                         X = 542 g of  CO2

Number of moles     542 g / 44.01 g/mol = 6.43 mol of MgCO3

d)                                       2C2H2 + 5O2 → 2H2O + 4CO2

Molar mass             C2H2 = 26.04 g/mol             H2O = 18.01528 g/molProportions              20.0 Kg of C2H2X g of H2O

                                         X = 13.8 Kg of  H2O

Number of moles     13.8 Kg / 18.01528 g/mol = 768 mol of H2O

e)                                       2BaO2 → 2BaO + O2

Molar mass             BaO2 = 169.33 g/mol             BaO = 153.33 g/molProportions              X g of BaO2        2.500 Kg of BaO

                                         X = 2.761 Kg of  BaO2

Number of moles     2.761 Kg / 169.33 g/mol = 16.30 mol of BaO2