What volume in milliliters of 0.0150 M Ba(OH)2 is required to neutralize 50.0 mL of 0.0500 M HBr?
1 answer:
The volume of required will be 83.33 mL
The equation of the reaction is as follows:
Mole ratio of reactants = 1:2
Mole of 50 mL, 0.05 M HBr = 0.05 x 0.05 = 0.0025 moles
Equivalent mole of = 0.0025/2 = 0.00125 moles
Volume of 0.015 M, 0.00125 moles = 0.00125/0.015 = 0.0833 L
0.0833 L = 83.33 mL
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Answer:
Hi, the question is incomplete. However, the question is about the calculation of volume of a product when the volume of one of the reactants is provided.
9.587 cm^3
Explanation:
The balanced equation for the chemical reaction is shown below:
⇒
In the chemical reaction above, 2 moles of water produced 4 moles of hydrogen fluoride. If 4.8 cm^3 of water were consumed, we can calculated the volume of hydrogen fluoride that would be produced as follow:
Using STP, 1 mole of gas has a volume of 22.4 L
Thus, 4.8 cm^3 = 0.0048 L is equivalent to 2.14*10^-4
since 2 moles of water produced 4 moles of hydrogen fluoride, therefore, 2.14*10^-4 would produced 2*2.14*10^-4 = 4.28*10^-4 moles
we can convert the moles to L by multiplying with 22.4
volume of hydrogen fluoride = 4.28*10^-4 * 22.4 = 0.009587 L = 9.587 cm^3