Limiting reagent problem calculate the theoretical yield of...

Limiting reagent problem: Calculate the theoretical yield of the precipitate for the precipitation of cobalt (II) carbonate (CoCO3). This question is part of a lab where metal nitrate is dissolved in water. Sodium carbonate solution is mixed with cobalt (II) nitrate. The solution is filtered and oven-dried on a gooch filter (mass of precipitate already has gooch filter mass subtracted from gooch+precipitate value). Balanced Equation: Na2CO3 + Co(NO3)2 -> CoCO3 (not soluble, solid) + 2NaNO3 (soluble) Mass of metal nitrate: 0.408g NaCl volume: 9.64mL Mass of precipitate: 0.071g Na2CO3 = 0.100 M

Transcript

00:01 Hello students, let's begin with this question.

00:02 In this question this reaction is given and the mass of the metal nitrate that means the mass of the co₂ nitrate is given which is equals to 0 .480 then the molarity of na₂co₃ is given which is equals to 0 .100 molar.

00:33 Then the volume of the na₂co₃ is given which is equals to 9 .64 ml.

00:42 So in this question we have to find out the theoretical yield of the cobalt two carbonate that is co₃co₃.

00:53 So now firstly we have to find out the moles here.

00:58 So firstly the moles of the cobalt nitrate is equals to its mass is given so mass of the co₃co₃ then divided by its molar mass.

01:12 So now by substituting the values in this formula we get here the moles.

01:18 So mass is equals to 0 .408 gram and its molar mass is equals to 206 .95 gram per mole.

01:30 So when we solve this we get here the moles of the cono₃₂ so which is equals to 0 .00197 moles.

01:42 Now then the molarity and the volume of the na₂co₃ is given.

01:47 So by using the formula that moles of na₂co₃ is equals to the molarity of na₂co₃ multiplied by the volume in liters.

02:04 So now substituting the values in this formula so molarity is equals to 0 .100 moles per liter multiplied by 9 .64 divided by 1000 liter.

02:18 So when we solve this we get here the so now the moles of the sodium carbonate comes to be 0 .000964 moles...

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