Methanol, ethanol, and n−propanol are three common alcohols. When 1.00 g of each of these alcohols is burned in air, heat is lib
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Answer: -
Volume of NaOH = 461 mL = 461/1000 = 0.461 L
pH = 10.0
Using the relation pH + pOH = 14.00
pOH = 14.00 - 10.00= 4.00
Using the relation
[OH⁻] =
= M
Since NaOH is a monobasic acid,
[NaOH] = [OH⁻] = M
Amount of NaOH = [NaOH] x Volume of NaOH
= M x 0.461 L
= 4.61 x 10⁻⁵ mol
Molar mass of NaOH = 23 x 1 + 16 x 1 + 1 x1
= 40 g / mol
Mass of NaOH required = 4.61 x 10⁻⁵ mol x 40 g/ mol
= 1.84 x 10⁻³ g
Answer:
Explanation:
Let's examine this reaction, especially the coefficients.
First of all, remember that reactants are used to make the product. They are found on the left of the arrow usually. Therefore, N₂ and H₂ are the reactants, while NH₃ is the product. We can automatically eliminate NH₃ from our answer choices, because it is simply not a reactant.
Next, look at the coefficients.
- N₂ has no coefficient, so a 1 is implied.
- H₂ has a coefficient of 3.
Therefore, for the reaction to work, there must be 1 mole of N₂ and 3 moles of H₂.
We have 3.2 moles of N₂ and 5.4 moles of H₂.
Divide each amount given by the required amount for completion.
- 3.2 mol N₂/ 1 mol N₂= 3.2 times
- 5.4 mol H₂/ 3 mol H₂=1.8 times
Therefore, there is enough nitrogen to complete the reaction 3.2 times, but only enough hydrogen for 1.8 times. If everything is completely reacted, we will run out of hydrogen and have excess <u>nitrogen</u>.