Answer:
0.0488 L
Explanation:
Step 1:
The balanced equation for the reaction. This is given below:
CaCl2(aq) + Na2CO3(aq) —> CaCO3(s) + 2NaCl(aq)
Step 2:
Determination of the number of mole of Na2CO3 in 0.0650 L of 1.50 M Na2CO3 solution.
Volume of solution = 0.0650 L
Molarity of Na2CO3 = 1.50 M
Mole of solute (Na2CO3) =?
Molarity = mole of solute /Volume of solution
1.50 = mole of solute/0.0650
Cross multiply to express in linear form.
Mole of solute = 1.5 x 0.0650
Mole of solute (Na2CO3) = 0.0975 mole
Step 3:
Determination of the number of CaCl2 that reacted.
CaCl2(aq) + Na2CO3(aq) —> CaCO3(s) + 2NaCl(aq)
From the balanced equation,
1 mole of CaCl2 reacted with 1 mole Na2CO3.
Therefore, 0.0975 mole of CaCl2 will also react with 0.0975 mole of Na2CO3.
Step 4:
Determination of the volume of CaCl2 that reacted.
Mole of solute (CaCl2) = 0.0975 mole
Molarity of CaCl2 = 2.00 M
Volume of solution =?
Molarity = mole of solute /Volume
2 = 0.0975/volume
Cross multiply to express in linear form
2 x Volume = 0.0975
Divide both side by 2
Volume = 0.0975/2
Volume = 0.0488 L
Therefore, the volume of CaCl2 that is 0.0488 L
