Posted by kelly on Monday, August 26, 2013 at 8:58pm.
How much solid product formed with 30ml of NaOH and 40mL of NiCl2?
I'm not really sure how to go about this?
incomplete---Chemistry - DrBob222, Monday, August 26, 2013 at 9:40pm
Can't be done. Need concentrations of NaOH and NiCl2.
Chemistry - kelly, Monday, August 26, 2013 at 10:20pm
They are both 1Mole
Chemistry - DrBob222, Monday, August 26, 2013 at 11:12pm
mole is not a concentration. You may mean 1 M which is one (1) molar.
2NaOH + NiCl2 ==> 2NaCl + Ni(OH)2
mols NaOH = M x L = 1M x 0.030 = 0.030
mols NiCl2 = 1M x 0.040 = 0.040
Now use the coefficients in the balanced equation to convert mols NaOH to mols Ni(OH)2 and mols NiCl2 to mols Ni(OH)2.
First, convert NaOH.
mols Ni(OH)2 = 0.03 mols NaOH x (1 mol Ni(OH)2/2 mols NaOH) = 0.03 x (1/2) = 0.015 mols Ni(OH)2.
Next, convert NiCl2.
mols Ni(OH)2 = 0.040 x (1 mol Ni(OH)2/1 mol NiCl2) = 0.04 x 1/1 = 0.04.
You have obtained two answers for mols Ni(OH)2 produced. Both can't be right; in limiting reagent problems (and this is one) the smaller answers is always the correct one and the reagent producing that value is the limiting reagent. Therefore, there will be 0.015 mol Ni(OH)2 produced. If you want grams covert that value. g = mols x molar mass.
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