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Na2CO3(aq) + CaCl2(aq) — 2 NaCl(aq) + CaCO3(s)
Calculate the volume (in mL) of 0.200 M CaCl, needed to produce 2.00 g of CaCO3(s). There is an excess of Na2CO3.
Molar mass of calcium carbonate = 100.09 g/mol

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Answer:

100 mL

Explanation:

Given data:

Mass of CaCO₃ produced = 2.00 g

Molarity of CaCl₂ = 0.200 M

Volume of CaCl₂ needed = ?

Solution:

Chemical equation:

Na₂CO₃  + CaCl₂    →      2NaCl + CaCO₃

First of all we will calculate the number of moles of CaCO₃.

Number of moles = mass/molar mass

Number of moles = 2.00 g / 100.09 g/mol

Number of moles = 0.02 mol

Now we will compare the moles of CaCO₃ and CaCl₂.

              CaCO₃          :            CaCl₂

                   1                :               1

                 0.02           :              0.02

Thus, 0.02 moles of CaCl₂ react,

Volume of CaCl₂ reacted:

Molarity = number of moles / volume in L

0.200 M = 0.02 mol / volume in L

Volume in L = 0.02 mol / 0.200 M

Volume in L = 0.1 L

Volume in mL:

0.1 L × 1000 mL/1L

100 mL

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