Determine the volume of methane (ch 4 ) gas needed to react completely with 0.660 l of o 2 gas to form methanol (ch 3 oh).

Balance Chemical Equation for this reaction is,

                           2 CH₄  +  O₂   →   2CH₃OH

According to this eq, 22.4 L (1 moles) of Oxygen requires 44.8 L (2 mole) CH₄ for complete reaction.
So, the volume of CH₄ required to consume 0.66 L of O₂ is calculated as,

 22.4 L O₂ required to consume  =  44.8 L CH₄
0.660 L O₂ will require                =  X L of CH₄

Solving for X,
                                                X  =  (44.8 L × 0.660 L) ÷ 22.4 L

                                                X  =  1.320 L of CH₄

Result:
            1.320 L of CH₄ gas is needed to react completely with 0.660 L of O₂ gas to form methanol (CH₃OH).

The activity series is shown in the attached picture. if the element is below the other, then you can displace the other. following the activity series as a guide, the reactions are as follows: nickel and aluminum chloride3 ni + 2 alcl₃  → 3 nicl₂ + 2 albarium and copper bromideba + cubr  →  no reaction because cu is less oxidizable than bapotassium nitrate and aluminum3 kno₃ + al  → al(no₃)₃ + 3 ksodium phosphate and zincnaso₄ + zn  → znso₄ + na