Precipitation occurs when the product of the ion concentration exceeds the Ksp.
M CH₃COOH: 12u×2 + 1u×4 + 16u×2 =<u> 60u</u>
m 9CH₃COOH: 60u×9 = <u>540u</u>
<em>(1u ≈ 1,66·10⁻²⁴g)</em>
-----------------------------
1u ------- <span>1,66·10⁻²⁴g
540u ---- X
X = 540</span>×<span>1,66·10⁻²⁴g
<u>X = 896,4</u></span><span><u>·10⁻²⁴g
</u></span>
Answer: 8.30 g of calcium sulfate are produced from 10 grams of lithium sulfate.
Explanation:
To calculate the moles :
According to stoichiometry :
1 mole of 
require = 1 mole of 
Thus 0.061 moles of will require=
of 
Thus is the limiting reagent as it limits the formation of product and is the excess reagent.
As 1 mole of give = 1 mole of 
Thus 0.061 moles of give = of
Mass of 
Thus 8.30 g of calcium sulfate are produced from 10 grams of lithium sulfate.
Answer:
The answer to your question is <u>111 g of CaCl₂</u>
Explanation:
Reaction
2HCl + CaCO₃ ⇒ CaCl₂ + CO₂ + H₂O
Process
1.- Calculate the molecular mass of Calcium carbonate and calcium chloride
CaCO₃ = (1 x 40) + (1 x 12) + ((16 x 3) = 100 g
CaCl₂ = (1 x 40) + (35.5 x 2) = 111 g
2.- Calculate the amount of calcium chloride produced using proportions.
The proportion CaCO₃ to CaCl₂ is 1 : 1.
100 g of CaCO₃ ------------- 111 g of CaCl₂
Then 111g of CaCl₂ will be produced.
