Answer:
The answer to your question is 1.1 moles of water
Explanation:
2Al(OH)₃ + 3H₂SO₄ ⇒ Al₂(SO₄)₃ + 6H₂O
0.45 mol 0.55 mol ?
Process
1.- Calculate the limiting reactant
Theoretical proportion
Al(OH)₃ / H₂SO₄ = 2/3 = 0.667
Experimental proportion
Al(OH)₃ / H₂SO₄ = 0.45 / 0.55 = 0.81
From the proportions, we conclude that the limiting reactant is H₂SO₄
2.- Calculate the moles of H₂O
3 moles of H₂SO₄ ---------------- 6 moles of water
0.55 moles of H₂SO₄ ----------- x
x = (0.55 x 6) / 3
x = 3.3 / 3
x = 1.1 moles of water
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It will received an H+. I’m not sure if that answers your question correctly
The volume of water that will be produced from the reaction will be 6.3 mL
<h3>Stoichiometric calculation</h3>
From the equation of the reaction:
The mole ratio of hydrogen sulfate to sodium hydroxide is 1:2.
Mole of hydrogen sulfate = 0.50 x 350/1000 = 0.175 moles
Mole of 15 grams sodium hydroxide = 15/40 = 0.375 moles
Thus, hydrogen sulfide is the limiting reagent.
Mole ratio of hydrogen sulfide to water = 1:2.
Equivalent mole of water = 0.175 x 2 = 0.35 moles
Mass of 0.35 moles of water = 0.35 x 18 = 6.3 grams.
1 gram of water = 1 ml.
Thus, 6.3 grams of water will be equivalent to 6.3 mL
More on stoichiometric calculation can be found here: brainly.com/question/27287858
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