Question

1. How many liters of oxygen at 23° C and 1.12 atm will be required for the combustion of 10.0 g of methane, CH₄, according to the chemical reaction below:

                                 CH₄ (g) + 2 O₂ (g)_­ ---> CO₂ (g) + 2 H₂O (g)  

2. When magnesium burns in air, the reaction is described by:  2 Mg (s) + O₂ (g) ---> 2 MgO (s)

What mass of magnesium will react with 10.8 L of oxygen if the temperature and pressure of the gas before the reaction are

-12° C and 664 mmHg?

3. Sulfur dioxide can be removed from other gases by its reaction with calcium oxide. The chemical equation for the reaction is

                                          CaO(s) + SO₂ (g) ----> CaSO₃ (s)

What mass of calcium oxide is required to react with 102 mL of sulfur dioxide gas at STP? STP is Standard Temperature (0° C) and Pressure (1 atm). PS: SO₂ is the yellow pollution emitted by the coal burning power plants (sulfur is naturally found in coal).

4. A mass of 0.0600 g of magnesium metal is completely reacted with excess hydrochloric acid according to:    

                                          Mg (s) + 2 HCl (aq) ----> H₂ (g) +MgCl₂ (aq)

What volume (in milliliters - mL) of hydrogen gas at STP would be produced? STP is Standard Temperature (0° C) and Pressure (1 atm).

Answer 1

Dear student,

Answer :- CHylg) + 202 (9) - CO2 (9) + 2 H₂OLI) e mole of Chy - ginen mass in molecular mass in g/mol 16 = 0.625 mol. for Imole of CH4 - 2 moles of ony gen is required for 0.625 mole of CHY – 2x0.625 moles of oxygen is required - 2x0.625 3 1.25 mol by ideal gas egen :- PN=nRT V = nRT where p = 1.12 atm 0.0821 lt-afm R = 0.0821 motok T= 23° = 23 + 273.15k = 296.15k. n = 1.25 mol. V = 1.25*0,0821 x 296.15 7:12 = 27.14 Lt

2 Mg(s) + O2 (9) 2190 189 a vol. of moles of O₂ gas in it 22.4 it by ideal gal eqation. Pra MRT P = 664 mmtig = 664 = 0,87 atm 760 V = 10.8 Lt = -12° c = 273.15 - 12K = 261.15 k. R = 0.0821 lt- atm mot-k PV =n mol n = 0.87 x 10,8 0.0821 X 261.15 = 0.44 mol Now, for a moles of O2 - 2 Moles of my es requered for 0.44 mole og 02 - 2x0.44 moles of Mg is required Mmg - 240.44 = 0.88 moles mass a magnisium a "mg x molar mass of my = 0.88 X 24 g = 21.12 g

cao (s) + O₂(g) - Caso3 (5) moles of S02 (9) = wol a gas in it 22.4 lt og gas at STP . ime = 103 Lt. moles = 102x10 3 2.2.4 - = 4.55x10 s moles - 4.55 mmol. for I moles z. SO2 (9) - I moles of caow is required for 4.55 mmol og so219) ↑ 4.55 mmel og caolss is required. caois) = 4.55 mmol. Mass & caolu = cao (s) & Molay mass of coolss. = 4.55 mmol x 56 g/mol = 254.8 mg -0.2548 g.

Mg 25) + 2HCl laq - H₂ (9) + Mg cla Caq) moles y Maes) = ginen mass in g molay mass in g/mol 0.06 24 = 0.0025 = 2.5 monol. for i moles of Mg (8) I roles of H₂29) is produce for 2.5 mmol of Mgls) 2.5 monol og H₂ (9) is produce mg = 2.5 mmol Now, at STP vol. f I moles gas = 22. y el at sap vol of Immol of gas = 2 at STP vol g 2.5 monol a gas = 22.4 x 2.5x103 lt = 22:4 x 10 vol og H₂19 = 22.4x2.5 x 10-3 et - 56X103 l { let = 105 mL. = 56ml

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