if 0.50 L of a 5.00 M stock solution of HCL is diluted to make 2.0 L of solution, how much HCL, in grams, is in the solution?
1 answer:
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The balanced molecular equation for the given reaction is:
To check that the equation is balanced count how many atoms are present for each element in the reactant and product side. If they are the same before and after reaction for all elements, then the reaction is deemed balanced.
The atom counting for the equation is shown below:
<u>Reactants </u> → <u>Products</u>
C: (2 x 2) =4 C: (2 x 2) = 4
H: (2 x 4) + (1 x 2) =10 H: (3 x 2) + (2 X 2) = 10
O: (2 x 2) + (1 x 2) = 6 O: (2 x 2) + (2 x 1) = 6
Ba: 1 Ba: 1
Since the number of atoms of each element are similar in the reactants and products, the equation is balanced.
To determine the state of the substances, consider their solubility.
The reactants are both aqueous (aq) as indicated in the problem.
The first product, is aqueous (aq) because based on the solubility rule, it will dissolve in water. Acetates are generally soluble.
The other product is water which will be liquid (l) since it is the solvent used to dissolve the substances.
<h3>LEARN MORE</h3>- Solubility Rules brainly.com/question/12984314
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Keywords: acid-base, neutralization, balancing equations, molecular equation
<u>Answer:</u> The amount of barium sulfate produced in the given reaction is 0.667 grams.
<u>Explanation:</u>
To calculate the number of moles from molarity, we use the equation:
Molarity of barium chloride = 0.113 M
Volume of barium chloride = 25.34 mL = 0.02534 L (Conversion factor: 1 L = 1000 mL)
Putting values in above equation, we get:
For the given chemical reaction:
By Stoichiometry of the reaction:
1 mole of barium chloride is producing 1 mole of barium sulfate.
So, 0.00286 moles of barium chloride will produce = of barium sulfate.
Now, to calculate the mass of barium sulfate, we use the equation:
Molar mass of barium sulfate = 233.38 g/mol
Moles of barium sulfate = 0.00286 moles
Putting values in above equation, we get:
Hence, the amount of barium sulfate produced in the given reaction is 0.667 grams