Step 1. Write and balance the equation. The one you have is incorrect. I have corrected it.
2Pb(NO3)2 ==> 2PbO + 4NO2 + O2
Step 2. Convert what you have to mols.
mols Pb(NO)2 = grams/molar mass = ?
Step 3. Convert mols of what you have to mols of what you want. Use the coefficients in the balanced equation to do that.
?mols Pb(NO3)2 x [2 mol PbO/2 mol Pb(NO)2] = ?
Step 4. Convert mols what you want (PbO here) to grams. g = mols x molar mass. This is the theoretical yield (TY). The actual yield (AY) is 9.4 g. These four steps will work all stoichiometry problems. Write them down. Memorize them.
Step 5. For % yield
% yield = (AY/TY)*100 = ?
The first problem is a limiting reagent problem. Calculate mols of PH formed as if EACH reactant were present by itself. That will give you two answers for mols PH3. The smaller one is the correct one to use. The remainder of the problem works just like problem 2. Post your work if you gt stuck.
8) ( 1 point) What is the maximum number of grams PH₃ that can be formed when 3.4 g of phosphorus reacts with 4 grams of hydrogen to form PH₃? P₄ (g) +6H₂ (g) → 4PH₃ (g) *
3.7 g
6.8 g
45 g
270 g
10. Lead nitrate can be decomposed by heating. What is the percent yield of the decomposition reaction if 15.6 g Pb(NO3)2 are heated to give 9.4 g of PbO? 2Pb(NO3)2(s)->PbO(s)+4NO2(g)+O2(g)
A. 59%
B. 73%
C. 82%
D. 90%
2 answers
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