The full ionic equation for the conversion of aqueous lead (II) nitrate to solid lead (II) sulfate and aqueous potassium nitrate is as follows:
<h3>What is
ionic equation ?</h3>
PbSO4(s) + 2 K+ + 2 NO3 = Pb2+(aq) + 2 NO3(aq), 2 K+(aq), and SO42(aq) (aq)
Let's look at the molecular formula that results in solid lead (II) sulfate and aqueous potassium nitrate when aqueous lead (II) nitrate and aqueous potassium sulfate interact. This reaction involves two displacements.
PbSO4(s) + 2 KNO3 = Pb(NO3)2(aq) + K2SO4(aq) (aq)
All ions and molecular species are included in the full ionic equation.
PbSO4(s) + 2 K+ + 2 NO3 = Pb2+(aq) + 2 NO3(aq), 2 K+(aq), and SO42(aq) (aq)
The full ionic equation for the conversion of aqueous lead (II) nitrate to solid lead (II) sulfate and aqueous potassium nitrate is as follows:
PbSO4(s) + 2 K+ + 2 NO3 = Pb2+(aq) + 2 NO3(aq), 2 K+(aq), and SO42(aq) (aq)
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Water, ALSO if you cant figure out the answer looking up on g00gl3e
the equation balanced is
3 PbI4 + 4 CrCl3 → 3 PbCl4 + 4 CrI3
Answer:
From the mole ratio of the gases, 10 moles of hydrogen will produce 10 moles of water, and if 10.0 L of hydrogen and 6.0 L of oxygen were used, 10.0 L of water will be produced.
What is the mole ratio of reactants in the formation of water?
The mole ratio of the reactants in the formation of water can be shown from the equation below:
Mole ratio of hydrogen to oxygen is 2:1
10 moles of hydrogen will require 5 moles of oxygen, therefore hydrogen is the limiting reactant.
10 moles of hydrogen will produce 10 moles of water.
1 mole of a gas has a volume of 22.4 L.
If 10.0 L of hydrogen and 6.0 L of oxygen were used, 10.0 L of water will be produced.
Therefore, the volume ratio of gas equals mole ratio of gas
Explanation: