What is the possible results of the test of a few drops of a mixture of sodium hydroxide and copper(ii) tetraoxosulphate (vi) so
1 answer:
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The limiting reagent will be Al
<h3>What are limiting reagents?</h3>They are reagents that limit the quantity of products that are formed in reactions.
From the equation of the reaction:
The mole ratio of Al to O2 is 4:3.
With 2 moles of Al and 2 moles of O2, Al becomes limiting while O2 is in excess.
With 2 moles of O2, the amount of Al required should be:
2 x 4/3 = 2.67 moles.
With 2 moles of Al, the amount of O2 required should be:
2 x 3/4 = 1.5 moles
Thus, O2 is in excess by 0.5 moles.
More on limiting reagents can be found here: brainly.com/question/11848702
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Answer:
A) 2.69 M
B) 0.059
Explanation:
A) We have:
33.8% solute by mass= 33.8 g solute/100 g solution
molarity = mol solute/ 1 L solution
molarity= x
x
x
molarity= 2.69 mol solute/L solution = 2.69 M
B) We know that there are 33.8 g of solute in 100 g of solution.
As the total solution is compounded by solute+solvent (in this case, solvent is water), the mass of water is the difference between the mass of the total solution and the mass of solute:
mass of water= 100 g - 33.8 g = 66.2 g
Now, we calculate the number of mol of both solute and water:
mol solute= 33.8 g solute x = 0.232 mol
mol H20= 66.2 g H₂O x
Finally, the mol fraction of solute (Xsolute) is calculated as follows:
Xsolute=
Xsolute= 0.059