How many liters of a 0.75 M solution of Ca(NO3)2 will be required to react with 148 g of Na2CO3? __ Ca(NO3)2 + __ Na2CO3 __ CaCO3 + __ NaNO3
Answers
Answer:
Explanation:
The given chemical reaction equation is:
Ca(NO₃)₂ + Na₂CO₃ --> CaCO₃ + 2 NaNO₃
Given mass of Na₂CO₃ = 148 grams
Molar mass of Na₂CO₃= 23 x 2 + 12 + 3 X 16 = 106 grams
Number of moles of Ca(NO₃)₂ = Given mass/ Molar mass
= 148/106
= 1.396 moles
From the chemical equation, 1 mole of Na₂CO₃ reacts with 1 mole of Ca(NO₃)₂.
⇒ 1.396 moles of Na₂CO₃ will react with 1.396 moles of Ca(NO₃)₂
So, We need to find how many liters of Ca(NO₃)₂ will contain 1.396 moles of Ca(NO₃)₂.
Strength of solution = 0.75 M
⇒ 0.75 moles of Ca(NO₃)₂ is present in 1 liter of solution.
⇒ 1.396 moles of Ca(NO₃)₂ is present in (1/0.75) x 1.396 liter of solution
⇒Volume of solution required = 0.95 liter