Answer:
Soluble salts can be made by reacting acids with soluble or insoluble reactants. Titration must be used if the reactants are soluble. Insoluble salts are made by precipitation reactions.
Making insoluble salts
An insoluble salt can be prepared by reacting two suitable solutions together to form a precipitate.
Determining suitable solutions
All nitrates and all sodium salts are soluble. This means a given precipitate XY can be produced by mixing together solutions of:
X nitrate
sodium Y
For example, to prepare a precipitate of calcium carbonate:
X = calcium and Y = carbonate
mix calcium nitrate solution and sodium carbonate solution together
calcium nitrate + sodium carbonate → sodium nitrate + calcium carbonate
Ca(NO3)2(aq) + Na2CO3(aq) → 2NaNO3(aq) + CaCO3(s)
It also works if potassium carbonate solution or ammonium carbonate solution is used instead of sodium carbonate solution. Remember that all common potassium and ammonium salts are soluble.
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Explanation:
Answer:
Mass = 8.46 g
Explanation:
Given data:
Mass of water produced = ?
Mass of glucose = 20 g
Mass of oxygen = 15 g
Solution:
Chemical equation:
C₆H₁₂O₆ + 6O₂ → 6H₂O + 6CO₂
Number of moles of glucose:
Number of moles = mass/molar mass
Number of moles = 20 g/ 180.16 g/mol
Number of moles = 0.11 mol
Number of moles of oxygen:
Number of moles = mass/molar mass
Number of moles = 15 g/ 32 g/mol
Number of moles = 0.47 mol
now we will compare the moles of water with oxygen and glucose.
C₆H₁₂O₆ : H₂O
1 : 6
0.11 : 6/1×0.11 = 0.66
O₂ : H₂O
6 : 6
0.47 : 0.47
Less number of moles of water are produced by oxygen thus it will limit the yield of water and act as limiting reactant.
Mass of water produced:
Mass = number of moles × molar mass
Mass = 0.47 mol ×18 g/mol
Mass = 8.46 g
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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