a) A reaction container holds 5.33 g of P4 and 3.77 g of O2 and reaction A occurs. If enough oxygen is available then the P4O6 reacts further to undergo reaction B. What is the limiting reactant for the formation of P4O6? b) What mass of P4O6 is produced (theoretical yield in grams)? c) If 7.12g of P4O6 were obtained what is the percent yield? d) Will reaction B occur? Why or why not? What mass of excess reactant is left in the reaction container?

The given question is incomplete, the complete question is:

Balance the following chemical equations: A) P4+ O2 → P406 B) _P406+ LO2 → P4010 a) A reaction container holds 5.33 g of P4 and 3.77 g of O2 and reaction A occurs. If enough oxygen is available then the P406 reacts further to undergo reaction B. What is the limiting reactant for the formation of P406? b) What mass of P406 is produced (theoretical yield in grams)? c) If 7.12g of P406 were obtained what is the percent yield? d) Will reaction B occur? Why or why not? What mass of excess reactant is left in the reaction container?

The balanced reaction will be,

A) P₄ + 3O₂ ⇒ P₄O₆

B) P₄O₆ + 2O₂ ⇒ P₄O₆

a) Based on the given information, the reaction container holds 5.33 grams of P₄ and 3.77 grams of oxygen. Thus, the moles of P₄ will be,

Moles = mass of P₄/Molar mass of P₄ = 5.33 grams/124 g/mole = 0.043 mole

Now the moles of O₂ will be,

Moles = mass of O₂/Molar mass of O₂ = 3.77 grams/32 g/mol = 0.112 mole

Now the moles of P₄O₆ formed when 0.043 moles of P₄ react completely will be = 1/1 × 0.043 = 0.043 mole of P₄O₆

Similarly, the moles of P₄O₆ formed, when 0.112 moles of O₂ react completely will be = 1/3 × 0.112 = 0.0373 mole of P₄O₆

Thus, from the analysis, the maximum moles of P₄O₆ formed will be 0.0373 moles. Therefore, oxygen will be the limiting reagent, which will react completely in the reaction.

b) From the above findings, the maximum moles of P₄O₆ produced is 0.0373 mole. Thus, the theoretical yield of P₄O₆ produced will be,

= Moles of P₄O₆ × Molar mass of P₄O₆

Theoretical yield = 0.0373 mole × 220 g/mole = 8.206 grams

c) Based on the given information, the actual mass of P₄O₆ produced is 7.12 grams.

Hence, percent yield = Actual yield/Theoretical yield * 100

= 7.12/8.206 × 100 = 86.77 %

d) In the given case, reaction B will not take place. This is due to the fact that oxygen is not left for reaction B, which was the limiting regent for reaction A. Here P₄ is the excess reactant, which was left in the reaction.

The initial moles of P₄ is 0.043, O₂ is 0.112, and P₄O₆ is O. The final moles of P₄ is 0.043 -1/3 × 0.112 = 0.0057 mole, O₂ is 0, and P₄O₆ is 0.0373 mole.

Thus, moles of P₄ left is 0.0057 mole. Hence, the mass of P₄ left will be,

= 0.0057 mole × Molar mass of P₄

= 0.0057 mole × 124 g/mole = 0.7068 grams.

uranium-238 i believe is your answer

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%