Answer:
[tex]V=100mL[/tex]
Explanation:
Hello.
In this case, since the chemical reaction is:
[tex]Na_2CO_3+CaCl_2\rightarrow CaCO_3+2NaCl[/tex]
We next compute the moles of sodium carbonate from the 2.00 grams of calcium carbonate via their 1:1 mole ratio in the chemical reaction:
[tex]n_{Na_2CO_3}=2.00gCaCO_3*\frac{1molCaCO_3}{100.09gCaCO_3}*\frac{1molNa_2CO_3}{1molCaCO_3} \\\\n_{Na_2CO_3}=0.0200molNa_2CO_3[/tex]
Thus, by knowing the molarity, we compute the volume:
[tex]M=\frac{n}{V}\\ \\V=\frac{n}{M}=\frac{0.0200mol}{0.200mol/L}\\ \\V=0.100L*\frac{1000mL}{1L}\\ \\V=100mL[/tex]
Best regards.
The volume in mL of 0.200 M Na₂CO₃ needed to produce 2.00 g of CaCO₃ there is an excess of CaCl₂ is 100mL.
Molarity is define as no. of moles of solute present in per liter of solvent or solution.
Given chemical reaction is:
Na₂CO₃ + CaCl₂ → CaCO₃ + 2NaCl
From the stoichiometry of the reaction it is clear that:
1 mole of CaCO₃ = produced by 1 mole of Na₂CO₃
Moles of CaCO₃ will be calculated as:
n = W/M, where
W = given mass = 2g
M = molar mass = 100.09g/mole
n = 2/100.09 = 0.0200 moles
0.0200 mole of CaCO₃ = produced by 0.0200 moles of Na₂CO₃
Now we calculate the volume by using molarity as:
V = n/M
V = 0.0200/0.2 = 0.1L = 100mL
Hence, the required volume is 100mL.
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