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
48.3 g AgNO3 / 169.9 g/mol = 0.284 moles AgNO3
0.284 mol AgNO3 X (1 mol Ag2CrO4/2 mol AgNO3) = 0.142 mol Ag2CrO4
0.142 mol Ag2CrO4 X 331.7 g/mol = 47.1 g Ag2CrO4
Answer:
Propane
Explanation:
From the question given, we were told that 0.1240 kg of propane reacted with excess oxygen to produce 0.3110kg of carbon dioxide.
Since the reaction took place in the presence of excess oxygen, therefore, propane is the limiting reactant as all of it is used up in the presence of excess oxygen.
To find the ratio of the the combination for the ion, write the charge of the cation as the subscript for the anion, and the charge of the anion as the subscript of the cation. This will make the charges effectively cancel and you will be left with a neutral ionic compound. Remember, that an ionic compound is made up of a metal and a nonmetal.
For Ca2+ and Cl-, you will get the neutral compound to be CaCl₂.
Answer:
17.55 g of NaCl
Explanation:
The following data were obtained from the question:
Molarity = 3 M
Volume = 100.0 mL
Mass of NaCl =..?
Next, we shall convert 100.0 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
100 mL = 100/1000
100 mL = 0.1 L
Therefore, 100 mL is equivalent to 0.1 L.
Next, we shall determine the number of mole NaCl in the solution. This can be obtained as follow:
Molarity = 3 M
Volume = 0.1 L
Mole of NaCl =?
Molarity = mole /Volume
3 = mole of NaCl /0.1
Cross multiply
Mole of NaCl = 3 × 0.1
Mole of NaCl = 0.3 mole
Finally, we determine the mass of NaCl required to prepare the solution as follow:
Mole of NaCl = 0.3 mole
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
Mass of NaCl =?
Mole = mass /Molar mass
0.3 = mass of NaCl /58.5
Cross multiply
Mass of NaCl = 0.3 × 58.5
Mass of NaCl = 17.55 g
Therefore, 17.55 g of NaCl is needed to prepare the solution.
