Answer:
100 mL of Na2CO3
Explanation:
We'll begin by calculating the number of mole in 1 g of CaCO3. This can be obtained as follow:
Mass of CaCO3 = 1 g
Molar mass of CaCO3 = 100.09 g/mol
Mole of CaCO3 =?
Mole = mass /Molar mass
Mole of CaCO3 = 1/100.09
Mole of CaCO3 = 0.01 mole
Next, we shall determine the number of mole of Na2CO3 needed to produce 0.01 mole of CaCO3.
This is illustrated below:
Na2CO3 + CaCl2 —> 2NaCl + CaCO3
From the balanced equation above,
1 mole of Na2CO3 reacted to produce 1 mole of CaCO3.
Therefore, 0.01 mole of Na2CO3 will also react to produce 0.01 mole of CaCO3.
Next, we shall determine the volume of Na2CO3 needed for the reaction as illustrated below:
Mole of Na2CO3 = 0.01 mole
Molarity of Na2CO3 = 0.1 M
Volume of Na2CO3 solution needed =?
Molarity = mole /Volume
0.1 = 0.01 / volume of Na2CO3
Cross multiply
0.1 × volume of Na2CO3 = 0.01
Divide both side by 0.1
Volume of Na2CO3 = 0.01 / 0.1
Volume of Na2CO3 = 0.1 L
Finally, we shall convert 0.1 L to millilitres (mL). This can be obtained as follow:
1 L = 1000 mL
Therefore,
0.1 L = 0.1 L × 1000 mL / 1 L
0.1 L = 100 mL
Thus, 0.1 L is equivalent to 100 mL.
Therefore, 100 mL of Na2CO3 is needed for the reaction.
