Answer:
CaCl2 (aq) + K2CO3(aq) ---------> CaCO3(s) + 2KCl(aq)
Explanation:
We have the reactants as calcium chloride and potassium carbonate. Recall that we are expecting that the reaction will yield a precipitate. We must keep that in mind as we seek to write its balanced chemical reaction equation.
So we now have;
CaCl2 (aq) + K2CO3(aq) ---------> CaCO3(s) + 2KCl(aq)
Recall that the rule of balancing chemical reaction equation states that the number of atoms of each element on the right side of the reaction equation must be the same as the number of atoms of the same element on the left hand side of the reaction equation.
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Answer:
17.55 g of NaCl
Explanation:
The following data were obtained from the question:
Molarity = 3 M
Volume = 100.0 mL
Mass of NaCl =..?
Next, we shall convert 100.0 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
100 mL = 100/1000
100 mL = 0.1 L
Therefore, 100 mL is equivalent to 0.1 L.
Next, we shall determine the number of mole NaCl in the solution. This can be obtained as follow:
Molarity = 3 M
Volume = 0.1 L
Mole of NaCl =?
Molarity = mole /Volume
3 = mole of NaCl /0.1
Cross multiply
Mole of NaCl = 3 × 0.1
Mole of NaCl = 0.3 mole
Finally, we determine the mass of NaCl required to prepare the solution as follow:
Mole of NaCl = 0.3 mole
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
Mass of NaCl =?
Mole = mass /Molar mass
0.3 = mass of NaCl /58.5
Cross multiply
Mass of NaCl = 0.3 × 58.5
Mass of NaCl = 17.55 g
Therefore, 17.55 g of NaCl is needed to prepare the solution.