TLDR: 6.53x10^5 g NH4ClO4
The stoichiometric coefficients (the numbers in front of the reactants and products) show that Aluminum and Ammonium Perchlorate are consumed at the exact same rate throughout the reaction: 3 parts of one to 3 parts of another.
1.5x10^5 grams of Aluminum, considering that the formula weight of Aluminum is 26.98 g/mol, is equal to 5,559.7 moles of Aluminum. This means that 5,559.7 moles of Ammonium Perchlorate are required to run the reaction to completion.
The formula weight of Ammonium Perchlorate is 117.49 grams a mole, and multiplying it by 5,559.7 moles to react to completion means that 6.53x10^5 grams of Ammonium Perchlorate is required for the reaction.
Answer:
a) Endothermic
b) T₂ = 53.1 ºC
Explanation:
a) We are told that when the ammonium nitrate dissolves in water the pack gets cold so the system is absorbing heat from the surroundings and by definition it is an endothermic process.
b) Recall that the heat, Q, is given by the formula:
Q = mcΔT where m is the mass of water,
c is the specific heat of water, and
ΔT is the change in temperature
We can determine the value for Q since we are given the heat of solution for the ammonium nitrate. From there we can calculate ΔT and finally answer our question.
Molar mass NH₄NO₃ = 80.04 g/mol
moles NH₄NO₃ = 50.0 g/ 80.04 g/mol = 0.62 mol
Q = 25.4 kJ/mol x 0.62 mol = 15.87 kJ = 15.87 kJ x 1000 J = 1.59 x 10⁴ J
Q = mcΔT ⇒ ΔT = Q/mc
ΔT = 1.59 x 10⁴ J/ (135 g x 4.184 J/gºC ) = 28.1 ºC
T₂- T₁ = ΔT ⇒ T₂ = ΔT + T₁ = 28.1 ºC +25.0 ºC = 53.1 ºC
Answer:
1640 kJ are involved in the reaction
Explanation:
In the reaction:
B₂H₆(g) + 6Cl₂(g) → 2 BCl₃(g) + 6HCl(g)
<em>1 mol of B₂H₆(g) with 6 moles of Cl₂(g) produce 1396 kJ of energy.</em>
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Now if 32.5g of B₂H₆(g) react with excess Cl₂(g), moles involved in reaction are:
If 1 mol produce 1396kJ of energy, 1.175 moles produce:
Thus, <em>1640 kJ are involved in the reaction</em>
