Answer:
10 molecules of NH₃.
Explanation:
N₂ + 3H₂ --> 2NH₃
As the N₂ supply is unlimited, what we need to do to solve this problem is <u>convert molecules of H₂ into molecules of NH₃</u>. To do so we use the <em>stoichiometric coefficients</em> of the balanced reaction:
- 15 molecules H₂ *
= 10 molecules NH₃
10 NH₃ molecules could be prepared from 15 molecules of H₂ and unlimited N₂.
Answer:
90%
Explanation:
Percentage yield = ?
Theoretical yield = 50g
Actual yield = 45g
To calculate the percentage yield of a compound, we'll have to use the formula of percentage yield which is the ratio between the actual yield to theoretical multiplied by 100
Percentage yield = (actual yield / theoretical yield) × 100
Percentage yield = (45 / 50) × 100
Percentage yield = 0.9 × 100
Percentage yield = 90%
The percentage yield of the substance is 90%
Answer:
49.95 g of HCl
Explanation:
Let's formulate the chemical equation involved in the process:
Ca(OH)2 + 2 HCl → CaCl2 + 2 H2O
This means that we need 1 mole of Calcium hydroxide to neutralize 2 moles of hydrochloric acid. From this, we calculate the quantity of HCl moles that would be neutralized by 0.685 moles of Ca(OH)2
1 mole Ca(OH)2 ---- 2 moles HCl
0.685 moles Ca(OH)2 ---- x = 1.37 moles HCl
Now that we know the quantity of HCl moles that would react, let's calculate the quantity of grams this moles represent:
1 mole of HCl ---- 36.46094 g
1.37 moles ------ x = 49.95 g of HCl
Two electrons that are of the same charge will repel each other
Answer:
At the cathode during the electrolysis of an aqueous solution of magnesium iodide, MgI2 , 2I−(aq) is produced
Explanation:
At cathode, reduction reaction takes place.
The dissociation of MgI2 in aqueous solution is Mg2+(aq) and 2I−(aq)
Here, the Iodine reduces to 2I−(aq) from state of 0 (MgI2) to state of -1 (2I−(aq))
Hence, at the cathode during the electrolysis of an aqueous solution of magnesium iodide, MgI2 , 2I−(aq) is produced
