The Ksp of calcium carbonate in water at 25 °C is 2.25 x 10-8. CaCO3(s) <----> Ca2+ (aq) + CO3 2- (aq) What is favored at
1 answer:
Answer:
the solubility of CaCO3 is 0.015g/l 25 °C
is favored at equilibrium
Explanation:
The Ksp of calcium carbonate in water at 25 °C is 2.25 x 10-8. CaCO3(s) <----> Ca2+ (aq) + CO3 2- (aq) What is favored at equilibrium?
solubility is the property of a solute to dissolve in a solvent(liquid, gas ) to form a solution(soution can be saturated ,unsaturated, or supersaturated)
CaCO3(s) <----> Ca2+ (aq) + CO3 2- (aq)
in partial dissociation , we can say
2.25x 10^-8=
let Ca^2+=CO3^-2=S
2.25x10^-8=S*S
S^2=2.25x10^-8
S=0.00015mol/L
Converting that to g/l
the relative molecular mass of CaCO3=100g/mol
0.00015*100g/mol
0.015g/l
the solubility of CaCO3 is 0.015g/l @room temperature
is favored at equilibrium
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Answer:
a. HCl.
b. 0.057 g.
c. 1.69 g.
d. 77 %.
Explanation:
Hello!
In this case, since the reaction between magnesium and hydrochloric acid is:
Whereas there is 1:2 mole ratio between them.
a) Here, we can identify the limiting reactant as that yielded the fewest moles of hydrogen gas product via the 1:1 and 2:1 mole ratios:
Thus, since hydrochloric yields fewer moles of hydrogen than magnesium, we realize it is the limiting reactant.
b) Here, we use the molar mass of gaseous hydrogen (2.02 g/mol) to compute the mass:
c) Here, we compute the mass of magnesium associated with the yielded 0.0248 moles of hydrogen:
Thus, the mass of excess magnesium turns out:
d) Finally, we compute the percent yield, considering 0.044 g is the actual yield and 0.057 g the theoretical yield:
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