The two main products of the combustion of gasoline in an automobile engine are
2 answers:
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Answer:
Explanation:
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In this case, we write the reaction again:
In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:
Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:
But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:
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Answer:
2.86mol/L
Explanation:
Given parameters:
Number of moles of MgCl₂ = 7.15moles
Volume of solution = 2.50L
Unknown:
Molarity of the MgCl₂ solution = ?
Solution:
The molarity of a solution is the number of moles of solute found in a given volume.
Molarity =
Insert the parameters and solve;
Molarity = = 2.86mol/L