Chemistry, asked by sruthypotter6061, 11 months ago

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

Answered by shashankvky
4

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

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