Answer:
The correct answer is 32.2 grams.
Explanation:
Based on the given information, the enthalpy of formation for aluminum oxide is 1676 kJ/mol. It signifies towards the energy that is required to generate aluminum and oxygen, and both of these exhibit zero enthalpy of formation. Therefore, the ΔHreaction is the required energy to generate 2 moles of aluminum. Thus, the energy needed for the formation of single mole of aluminum is,
ΔHrxn = 1676/2 = 838 kJ/mol
Q or the energy input mentioned in the given case is 1000 kJ. Therefore, the number of moles of Al generated is,
(1000 kJ) / (838 kJ/Al mole) = 1.19 moles of Aluminum
The grams of aluminum produced can be obtained by using the formula,
mass = moles * molecular mass
= 1.19 * 26.98
= 32.2 grams.
Answer:
The answer to your question is 41.6 g of AgCl
Explanation:
Data
mass of NH₄Cl = 15.5 g
mass of AgNO₃ = excess
mass of AgCl = 35.5 g
theoretical yield = ?
Process
1.- Write the balanced chemical reaction.
NH₄Cl + AgNO₃ ⇒ AgCl + NH₄NO₃
2.- Calculate the molar mass of NH₄Cl and AgCl
NH₄Cl = 14 + 4 + 35.5 = 53.5 g
AgCl = 108 + 35.5 = 143.5 g
3.- Calculate the theoretical yield
53.5 g of NH₄Cl -------------------- 143.5 g of AgCl
15.5 g of NH₄Cl ------------------- x
x = (15.5 x 143.5) / 53.5
x = 2224.25 / 53.5
x = 41.6 g of AgCl
