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It has more surface area to react with
Sodium citrate is obtained by mixing citric acid with hydroxide or some other sodium salt.so sodium citrate is by definition is a type of salt
1. You could drain off the oil as a supernatant, since oil is less dense than water.
2. You could attempt to evaporate the water since they have different boiling points (212 vs 572 F), and then condense it in a separate container using a distillation unit.
The pounds of alum produced when 0.26 g of hydrogen was produced are 0.0434 lb.
First, let's convert 0.126 g of hydrogen to moles using its molar mass (2.02 g/mol).
Let's consider the steps to make alum (KAl(SO₄)₂⋅12H₂O) from aluminum (Al).
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2 Al(s) + 2 KOH(aq) + 6 H₂O(l) →2 KAl(OH)₄(aq) + 3 H₂(g)
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2 KAl(OH)₄(aq) + H₂SO₄(aq) → 2 Al(OH)₃(s) + K₂SO₄(aq) + 2 H₂O(l)
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2 Al(OH)₃(s) + H₂SO₄(aq) → Al₂(SO₄)₃(aq) + 6 H₂O(l)
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K₂SO₄(aq) + Al₂(SO₄)₃(aq) + 24 H₂O(l) → 2 KAl(SO₄)₂⋅12H₂O(s)
To relate H₂ to KAl(SO₄)₂⋅12H₂O, we need to consider the appropriate molar ratios:
- In step 1, the molar ratio of H₂ to KAl(OH)₄ is 3:2.
- In step 2, the molar ratio of KAl(OH)₄ to Al(OH)₃ is 2:2.
- In step 3, the molar ratio of Al(OH)₃ to Al₂(SO₄)₃ is 2:1.
- In step 4, the molar ratio of Al₂(SO₄)₃ to KAl(SO₄)₂⋅12H₂O is 1:2.
The moles of KAl(SO₄)₂⋅12H₂O produced from 0.0624 moles of H₂ are:
The molar mass of alum is 474.38 g/mol. The mass corresponding to 0.0416 moles is:
Finally, we convert 19.7 grams to pounds using the conversion factor 1 lb = 454 g.
The pounds of alum produced when 0.26 g of hydrogen was produced are 0.0434 lb.
You can learn more about molar ratios here: brainly.com/question/15973092
Hey there!:
Write the molecular equation for the reaction of MgSO4 with Pb(NO3)2 :
MgSO4(aq) + Pb(NO3)2(aq) ---> Mg(NO3)2(aq) + PbSO4(s)
Write the total ionic equation for the reaction :
Mg²⁺ (aq) + SO₄⁻² (aq) + Pb²⁺ (aq) + 2 NO₃⁻¹ (aq) + PbSO₄(s)
Therefore:
Cancel the spectator ions on both sides:
Pb²⁺ (aq) + SO₄⁻² (aq) ---> PbSO4(s)
Hope that helps!