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%
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.
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To solve these problems, we need to use stoichiometry and the concept of limiting reactants. Let's break down each problem step by step.
8) To find the maximum number of grams of PH₃ that can be formed, we need to determine the limiting reactant first. The limiting reactant is the reactant that is completely consumed, limiting the amount of product formed.
1. Calculate the number of moles of each reactant:
Moles of phosphorus (P₄) = mass of phosphorus (3.4 g) / molar mass of P₄
Moles of hydrogen (H₂) = mass of hydrogen (4 g) / molar mass of H₂
2. Use the balanced equation to determine the stoichiometric ratio between P₄ and PH₃. From the balanced equation, we see that 1 mole of P₄ produces 4 moles of PH₃.
3. Calculate the number of moles of PH₃ that can be formed from each reactant:
Moles of PH₃ from P₄ = Moles of P₄ * (4 moles of PH₃ / 1 mole of P₄)
Moles of PH₃ from H₂ = Moles of H₂ * (4 moles of PH₃ / 6 moles of H₂)
4. Compare the moles of PH₃ calculated from each reactant. The smaller value will be the limiting reactant.
5. Finally, calculate the mass of PH₃ that can be formed from the limiting reactant:
Mass of PH₃ = Moles of PH₃ from the limiting reactant * molar mass of PH₃
10) To find the percent yield of the decomposition reaction, we need to compare the actual yield (9.4 g) with the theoretical yield (the maximum amount of product that can be formed).
1. Start by calculating the number of moles of Pb(NO₃)₂ using its mass and molar mass.
2. Using the balanced equation, determine the stoichiometric ratio between Pb(NO₃)₂ and PbO. From the balanced equation, we can see that 2 moles of Pb(NO₃)₂ produce 1 mole of PbO.
3. Calculate the theoretical yield of PbO by multiplying the number of moles of Pb(NO₃)₂ by the stoichiometric ratio.
4. Finally, calculate the percent yield by dividing the actual yield (9.4 g) by the theoretical yield and multiplying by 100%.
Once you have completed these steps, you will have the answers to the questions.