Answer:
Option A.
2Na + 2H2O —> 2NaOH + H2
Explanation:
To know which option is correct, we shall do a head count of the number of atoms present on both side to see which of them is balanced. This is illustrated below below:
For Option A:
2Na + 2H2O —> 2NaOH + H2
Reactant >>>>>>> Product
2 Na >>>>>>>>>>> 2 Na
4 H >>>>>>>>>>>> 4 H
2 O >>>>>>>>>>>> 2 O
Thus, the above equation is balanced.
For Option B:
2Na + 2H2O —> NaOH + H2
Reactant >>>>>>> Product
2 Na >>>>>>>>>>> 1 Na
4 H >>>>>>>>>>>> 3 H
2 O >>>>>>>>>>>> 1 O
Thus, the above equation is not balanced.
For Option C:
2Na + H2O —> 2NaOH + H2
Reactant >>>>>>> Product
2 Na >>>>>>>>>>> 2 Na
2 H >>>>>>>>>>>> 4 H
1 O >>>>>>>>>>>> 2 O
Thus, the above equation is not balanced.
For Option D:
Na + 2H2O —> NaOH + 2H2
Reactant >>>>>>> Product
1 Na >>>>>>>>>>> 1 Na
4 H >>>>>>>>>>>> 5 H
2 O >>>>>>>>>>>> 1 O
Thus, the above equation is not balanced.
From the illustrations made above, only option A is balanced.
From the stoichiometry of the combustion reaction, we can see that 7.4 L of oxygen is consumed.
<h3>What is combustion?</h3>
Combustion is a reaction in which a substance is burnt in oxygen. The equation of the reaction is; C4H10O(l) + 6O2 (g) → 4CO2 (g) + 5H2O(l)
We can obtain the number of moles of CO2 from;
PV = nRT
n = 1.02 atm * 7.15 L/0.082 atm LK-1mol-1 * (125 + 273) K
n = 7.29 /32.6
n = 0.22 moles
If 6 moles of oxygen produces 4 moles of CO2
x moles of oxygen produces 0.22 moles of CO2
x = 0.33 moles
1 mole of oxygen occupies 22.4 L
0.33 moles of oxygen occupies 0.33 moles * 22.4 L/ 1 mole
= 7.4 L of oxygen
Learn more about stoichiometry: brainly.com/question/13110055
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Yeah what the other person said ^
Answer:
Q = 10.8 KJ
Explanation:
Given data:
Mass of Al= 100g
Initial temperature = 30°C
Final temperature = 150°C
Heat required = ?
Solution:
Specific heat of Al = 0.90 j/g.°C
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 150°C - 30°C
ΔT = 120°C
Q = 100g×0.90 J/g.°C× 120°C
Q = 10800 J (10800j×1KJ/1000 j)
Q = 10.8 KJ
