The volume of water that will be produced from the reaction will be 6.3 mL
<h3>Stoichiometric calculation</h3>
From the equation of the reaction:
The mole ratio of hydrogen sulfate to sodium hydroxide is 1:2.
Mole of hydrogen sulfate = 0.50 x 350/1000 = 0.175 moles
Mole of 15 grams sodium hydroxide = 15/40 = 0.375 moles
Thus, hydrogen sulfide is the limiting reagent.
Mole ratio of hydrogen sulfide to water = 1:2.
Equivalent mole of water = 0.175 x 2 = 0.35 moles
Mass of 0.35 moles of water = 0.35 x 18 = 6.3 grams.
1 gram of water = 1 ml.
Thus, 6.3 grams of water will be equivalent to 6.3 mL
More on stoichiometric calculation can be found here: brainly.com/question/27287858
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Answer:
[C] = 0.4248M
Explanation:
A + B ⇄ 2C
C(i) 1.68M 1.68M 0.00
ΔC -x -x +2x
C(eq) 1.68-x 1.68-x 2x
Keq = [C]²/[A][B] = (2x)²/(1.68 - x)²= 8.98 x 10⁻²
Take SqrRt of both sides => √(2x)²/(1.68 - x)² = √8.98 x 10⁻²
=> 2x/1.68 - x = 0.2895
=> 2x = 0.2895(1.68 - x)
=> 2x = 0.4863 - 0.2895x
=> 2x + 0.2895x = 0.4863
=> 2.2895x = 0.4863
=> x = 0.4863/2.2895 = 0.2124
[C] = 2x = 2(0.2124)M = 0.4248M in 'C'
Answer:
Ksp = [Ca⁺²] × [C₂O₄⁻²]
Explanation:
Step 1: Write the balanced reaction for the dissociation of calcium oxalate
CaC₂O₄(s) ⇄ Ca⁺²(aq) + C₂O₄⁻²(aq)
Step 2: Write the expression for the solubility product constant (Ksp) of calcium oxalate
The solubility product constant is the equilibrium constant for the dissociation reaction, that is, it is equal to the product of the concentrations of the products raised to their stoichiometric coefficients divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients. It doesn't include solids nor pure liquids because their activities are 1.
Ksp = [Ca⁺²] × [C₂O₄⁻²]
That state of matter is known as liquid.
