If 0.500 mole of calcium carbonate is produced in a particular reaction, how many moles of sodium chloride are produced in the same reaction?

CaCl2 (aq) + Na2CO3 (aq) → CaCO3 (s) + 2 NaCl (aq)

Answer: 1mole of NaCl

Explanation:

Looking at the given equation : CaCl2 (aq) + Na2CO3 (aq) → CaCO3 (s) + 2 NaCl (aq)

The following deductions can be made,

1 mole of CaCl2 reacted with 1 mole of Na2CO3 to produce 1mole of CaCO3 and 2NaCl.

If 0.500 mole of CaCO3 was produced from this same reaction, it therefore means that we multiply the coefficients of both reactants and product by 0.500 to know the moles of NaCl that will also come out from this same reaction ...

So we have the new equation to be:

0.5CaCl2 (aq) + 0.5 Na2CO3 (aq) → 0.5 CaCO3 (s) + 1 NaCl (aq)

From the new equation, 1mole of NaCl will be produce alongside 0.5mole of CaCO3, when 0.5mole of CaCl2 reacts with. 0.5mole Na2CO3.

In the same reaction, 1.00 moles of NaCl are produced

Explanation:

Step 1: Data given

Number of moles calcium carbonate = 0.500 moles

Step 2: The balanced equation

CaCl2 (aq) + Na2CO3 (aq) → CaCO3 (s) + 2 NaCl (aq)

Step 3: Calculate moles sodium chloride (NaCl)

For 1 mol CaCl2 we need 1 mol Na2CO3 to produce 1 mol CaCO3 and 2 moles NaCl

For 0.500 moles CaCO3 produced we we'll have 2 * 0.500 = 1.00 mol of NaCl produced

In the same reaction, 1.00 moles of NaCl are produced