Answer:
To release 7563 kJ of heat, we need to burn 163.17 grams of propane
Explanation:
<u>Step 1</u>: Data given
C3H8 + 5O2 -----------> 3CO2 + 4H2O ΔH° = –2044 kJ
This means every mole C3H8
Every mole of C3H8 produces 2044 kJ of heat when it burns (ΔH° is negative because it's an exothermic reaction)
<u>Step 2: </u>Calculate the number of moles to produce 7563 kJ of heat
1 mol = 2044 kJ
x mol = 7563 kJ
x = 7563/2044 = 3.70 moles
To produce 7563 kJ of heat we have to burn 3.70 moles of C3H8
<u>Step 3: </u>Calculate mass of propane
Mass propane = moles * Molar mass
Mass propane = 3.70 moles * 44.1 g/mol
Mass propane = 163.17 grams
To release 7563 kJ of heat, we need to burn 163.17 grams of propane
Answer:
F2 is the limiting reactant
27.6 grams of NaF is produced.
Explanation:
Balance the equation first.
2Na+ F2 ---> 2NaF
To find the limiting reactant, solve for how much NaF can be produced with Na and F2
12.5g F2 x (1 mole F2/ 38.00 grams F2)x (2 mole NaF/ 1 mole F2)
=0.658 moles NaF
16.2g Na x (1 mole Na/ 22.99 grams Na)x (2 mole NaF/ 2 mole Na)
=0.705 moles NaF
Since F2 produced the least NaF, F2 is the limiting reactant.
Now, to find how much NaF there is, use the moles solved above with F2 as the limiting reactant.
0.658 moles NaF x (41.99 grams NaF/ 1 mole NaF)= 27.6 moles NaF
27.6 moles of NaF would be theoretically produced.
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<u>Explanation:</u>
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Atomic mass W = 183.84 u.m.a
183.84 g ----------- 6.02x10²⁴ atoms
?? g ---------------- 2.1x10²⁴ atoms
2.1x10²⁴ x 183.84 / 6.02x10²⁴ =
3.860x10²⁶ / 6.02x10²⁴ = 641.30 g
hope this helps!
Answer:
F<S<Na<Ga<Sr
Explanation:
Atomic radius increased from the right of the periodic table to left, and from the top to the down