The end product will depend upon
a) the amount of the reagent taken
b) the final treatment of the reaction
If we have just taken methylmagnesium iodide and p-hydroxyacetophenone, then we will get methane and hydroxyl group substituted with MgI in place of hydrogen
Figure 1
However if we have taken excess of methylmagnesium iodide which is Grignard's reagent followed by hydrolysis we will get different product
Figure 2
<u>Answer:</u> The metal having molar mass equal to 26.95 g/mol is Aluminium
<u>Explanation:</u>
.....(1)
Molarity of NaOH solution = 0.5000 M
Volume of solution = 0.03340 L
Putting values in equation 1, we get:
By Stoichiometry of the reaction:
2 moles of NaOH reacts with 1 mole of sulfuric acid
So, 0.01670 moles of NaOH will react with = of sulfuric acid
Excess moles of sulfuric acid = 0.00835 moles
Molarity of sulfuric acid solution = 0.5000 M
Volume of solution = 127.9 mL = 0.1279 L (Conversion factor: 1 L = 1000 mL)
Putting values in equation 1, we get:
Number of moles of sulfuric acid reacted = 0.06395 - 0.00835 = 0.0556 moles
By Stoichiometry of the reaction:
3 moles of sulfuric acid reacts with 2 moles of metal
So, 0.0556 moles of sulfuric acid will react with = of metal
Mass of metal = 1.00 g
Moles of metal = 0.0371 moles
Putting values in above equation, we get:
Hence, the metal having molar mass equal to 26.95 g/mol is Aluminium
Answer:
1.36 × 10³ mL of water.
Explanation:
We can utilize the dilution equation. Recall that:
Where <em>M</em> represents molarity and <em>V</em> represents volume.
Let the initial concentration and unknown volume be <em>M</em>₁ and <em>V</em>₁, respectively. Let the final concentration and required volume be <em>M</em>₂ and <em>V</em>₂, respectively. Solve for <em>V</em>₁:
Therefore, we can begin with 0.640 L of the 2.50 M solution and add enough distilled water to dilute the solution to 2.00 L. The required amount of water is thus:
Convert this value to mL:
Therefore, about 1.36 × 10³ mL of water need to be added to the 2.50 M solution.