Determine the mass (in g) of NiCO 3 that is produced when 770 mL of a 4.71×10 -2

Question

Determine the mass (in g) of NiCO3 that is produced when 770 mL of a 4.71×10-2 M Na2CO3 solution completely reacts with 626 mL of a 6.22×10-2 M NiCl2 solution according to the following balanced chemical equation.

Na2CO3(aq) + NiCl2(aq) NiCO3(s) + 2NaCl(aq)

ive gotten a mass of 0.317g but that is incorrect.

my steps were changing the molarity to moles my mulptlying by their respective volumes and found out wich was the limiting reagants. i then took that mole and mulptied it by the molar mass of NiCO3.

can someone tell me what i am doing incorrectly and help me get the right answer

Explanation / Answer

LEt's write the equation again:

Na2CO3(aq) + NiCl2(aq) NiCO3(s) + 2NaCl(aq)

Now, according to your data, you need to calculate the limiting reactant, this would be the one that will consume completely in the reaction and then, it will produce the mass of NiCO3 that you are looking for.

Let's first calculate the moles of both reactants, as we have a 1:1 relation of both of them, the one with the lowest mole will be the limiting reactant:

moles Na2CO3 = 0.0471 * 0.77 = 0.0362 moles
moles NiCl2 = 0.0622 * 0.626 = 0.0389 moles

The moles of Na2CO3 are the lowest, therefore, this is the limiting reactant. Now let's calculate the mass of NiCO3:
m NiCO3 = 0.0362 * 118.7023 = 4.297 g

This should be the correct answer. I don't know how you got the 0.317 g, but this should be right. Check that now.

Hope this helps

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