Answers
Answer:
1.17 L of H₂
Explanation:
We'll begin by calculating the number of mole in 2.3 g of Mg. This can be obtained as follow:
Mass of Mg = 2.3 g
Molar mass of Mg = 24 g/mol
Mole of Mg =?
Mole = mass /molar mass
Mole of Mg = 2.3 / 24
Mole of Mg = 0.096 mole
Next, we shall determine the number of mole of H₂ produced by the reaction of 2.3 g (i.e 0.096 mole) of Mg. This can be obtained as follow:
Mg + 2HCl —> MgCl₂ + H₂
From the balanced equation above,
1 mole of Mg reacted to 1 mole of H₂.
Therefore, 0.096 mole of Mg will also react to produce 0.096 mole of H₂.
Finally, we shall determine volume of H₂ produced from the reaction. This can be obtained as follow:
Number of mole (n) of H₂ = 0.096 mole
Pressure (P) = 2 atm
Temperature (T) = 298 K
Gas constant (R) = 0.0821 atm.L/Kmol
Volume (V) of H₂ =?
PV = nRT
2 × V = 0.096 × 0.0821 × 298
Divide both side by 2
V = (0.096 × 0.0821 × 298) /2
V = 1.17 L
Therefore, 1.17 L of H₂ were obtained from the reaction.
Related Questions
A beaker is filled to the 500 mL mark with alcohol. What increase in volume (in mL) does the beaker contain when the temperature changes from 5° C to 30° C? (Neglect the expansion of the beaker, evaporation of alcohol and absorption of water vapor by alcohol.) The volume coefficient of expansion γγ for alcohol = 1.12 x 10-4 K-1
Answers
Answer:
"1.4 mL" is the appropriate solution.
Explanation:
According to the question,
[tex]v_0=500[/tex][tex]\alpha =1.12\times 10^{-4}[/tex][tex]\Delta \epsilon = 25[/tex]Now,
Increase in volume will be:
⇒ [tex]\Delta V = \alpha\times v_0\times \Delta \epsilon[/tex]
By putting the given values, we get
[tex]=1.12\times 10^{-4}\times 500\times 25[/tex]
[tex]=1.12\times 10^{-4}\times 12500[/tex]
[tex]=1.4 \ mL[/tex]
A major component of gasoline is octane when octane is burned in air it chemically reacts with oxygen to produce carbon dioxide and water what mass of carbon dioxide is produced by the reaction of oxygen
Answers
gasoline is the chemical that is coming out of the air
