using the equaation 2h2+o2-->2h2o if 10.0g of hydrogen are used in the presence of excess oxygen how many grams of water will
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<u>Answer:</u> 0.0237 g of calcium carbonate would be required to neutralize the given amount of HCl
<u>Explanation:</u>
pH is defined as the negative logarithm of hydrogen ion concentration present in the solution
.....(1)
Given value of pH = 1.5
Putting values in equation 1:
Molarity is defined as the amount of solute expressed in the number of moles present per liter of solution. The units of molarity are mol/L. The formula used to calculate molarity:
.....(2)
We are given:
Volume of solution = 15.0 mL
Molarity of HCl = 0.0316 M
Putting values in equation 2:
The chemical equation for the reaction of HCl and calcium carbonate follows:
By the stoichiometry of the reaction:
2 moles of HCl reacts with 1 mole of calcium carbonate
So, of HCl will react with =
of calcium carbonate
The number of moles is defined as the ratio of the mass of a substance to its molar mass.
Moles of calcium carbonate =
Molar mass of calcium carbonate = 100.01 g/mol
Putting values in the above equation:
Hence, 0.0237 g of calcium carbonate would be required to neutralize the given amount of HCl
Answer : The concentration of and
at equilibrium is, 0.0158 M and 0.00302 M respectively.
Explanation :
First we have to calculate the concentration of
Now we have to calculate the value of equilibrium constant (K).
The given chemical reaction is:
Initial conc. 0.0163 0.00415 0.00276
At eqm. (0.0163-2x) (0.00415+x) (0.00276+x)
As we are given:
Concentration of at equilibrium = 0.00467 M
That means,
(0.00415+x) = 0.00467
x = 0.00026 M
Concentration of at equilibrium = (0.0163-2x) = (0.0163-2(0.00026)) = 0.0158 M
Concentration of at equilibrium = (0.00276+x) = (0.00276+0.00026) = 0.00302 M