When 2.00 L of 1.00 M of $A g N O_{3}$ $AgNO_3$ and 1.25 L of 1.00 M of $C a I_{2}$ $CaI_2$ fully react, what is the molarity of $C a^{2} +$ $Ca^2+$ ?

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When 2.00 L of 1.00 M of $A g N O_{3}$ $AgNO_3$ and 1.25 L of 1.00 M of $C a I_{2}$ $CaI_2$ fully react, what is the molarity of $C a^{2} +$ $Ca^2+$ ?

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Answer 1 · 2017-06-24T19:35:20

For $2 A g N O_{3} (a q) + C a I_{2} (a q) \to 2 A g I (s) ⏐ ↓ + C a {(N O_{3})}_{2} (a q)$ $2AgNO_3(aq) + CaI_2(aq) rarr 2AgI(s)darr + Ca(NO_3)_2(aq)$

Explanation:

The calcium ion, $C a^{2 +}$ $Ca^(2+)$ , which is present in solution as (probably) ${[C a {(O H_{2})}_{6}]}_{6}^{2 +}$ $[Ca(OH_2)_6]^(2+)$ , remains in solution, and expresses its concentration fully because it does not precipitate....

$Moles of$ $"Moles of"$ $C a^{2 +} = 1.25 \cdot L \times 1.00 \cdot m o l \cdot L^{- 1} = 1.25 \cdot m o l$ $Ca^(2+)=1.25*Lxx1.00*mol*L^-1=1.25*mol$ .

The volume of solution is $(2.00 + 1.25) \cdot L = 3.25 \cdot L$ $(2.00+1.25)*L=3.25*L$

And thus concentration of calcium ion is............

$(1.25*mol)/((2.00+1.25)*L)=0.385*mol*L^-1......$

Of course, the silver ion precipitates as a fine yellow powder of $AgI$ .........

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When 2.00 L of 1.00 M of $A g N O_{3}$ $AgNO_3$ and 1.25 L of 1.00 M of $C a I_{2}$ $CaI_2$ fully react, what is the molarity of $C a^{2} +$ $Ca^2+$ ?

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When 2.00 L of 1.00 M of AgNO_3AgNO3AgNO_3 and 1.25 L of 1.00 M of CaI_2CaI2CaI_2 fully react, what is the molarity of Ca^2+Ca2+Ca^2+?

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When 2.00 L of 1.00 M of $A g N O_{3}$ $AgNO_3$ and 1.25 L of 1.00 M of $C a I_{2}$ $CaI_2$ fully react, what is the molarity of $C a^{2} +$ $Ca^2+$ ?