#2 is easy. There is no reaction.
#1. This is a limiting reagent (LR) problem. You know that because amounts are given for both reactants.
First you need to balance the equation.
2CuNO3 + MgCl2 ---> 2CuCl + Mg(NO3)2
mols CuNO3 = grams/molar mass = ?
mols MgCl2 = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols CuNO3 to mols CuCl.
Do the same and convert mols MgCl2 to mols CuCl.
It is likely that the value for mols CuCl will not be the same; the correct value in LR problem is ALWAYS the smaller value and the reagent responsible for that number is the LR. That gives you the answer for 2. The excess reagent is the other one and that's the answer for 1.For mass of the excess reagent, isn't that just 2 g (or did you mean something else)?
For 3, mass CuCl = mols CuCl x molar mass CuCl. This is the theoretical yield (TY)
For the other product, use the LR, convert to mols Mg(NO3)22 formed and convert to g Mg(NO3)2.
4. The actual yield (AY) is given in the problem as 3.50 g CuCl.
%yield = (AY/TY)*100 = ?
I'm not sure how to do or show the work for these two questions. A) & B.)!! please anyone help:
A.) Suppose a chemical reaction starts with 5.00 g of copper (i) nitrate and 2.00 g of magnesium chloride.
Balanced Equation:
2 CuNo3 + MgCl2 ---> 2 CuCl + MgNO3
1.) determine the identity and mass of the excess reactant.
2.) Determine the identity of the limiting reactant.
3.) Determine the mass of each product that can be produced.
4.) Calculate the percent yield of copper (i) chloride if 3.50 g of copper (i) chloride is actually collected.
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B.) Suppose a chemical reaction starts with 15.0 g of sodium nitrate and 13.1 g of potassium sulfate.
1.) determine the identity & mass of excess reactant.
2.) Determine identity of limiting reactant
3.) Determine mass of each product that can be produced.
4.) Calculate percent yiled of potassium nitrate if 11.9 g of potassium nitrate is actually collected.
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1 answer