2 C5H12O + 15 O2 = 10 CO2 + 12 H2O
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This answer to this question is a rule that is applied to any reaction taken at dynamic equilibrium, with respect to 500 K. In other words, you can say that this reaction is of no use to us -
In a chemical equilibrium, it is known that the forward and reverse reactions occur at equal rates. At this point the concentrations of products and reactants remain constant, or in other words do not change
<u><em>Solution = Option C</em></u>
Answer:
171.34 g/mol
Explanation:
Ba molar mass = 137.328 g/mol
O molar mass = 15.999 g/mol * 2 = 31.9980 g/mol
H molar mass = 1.008 g/mol * 2 = 2.0160 g/mol
137.328 + 31.9980 + 2.0160 = 171.3420 = 171.34 g/mol
Answer:
Explanation:
Hello,
By assuming STP conditions (0°C and 1atm), we first compute the reacting moles of both hydrogen and nitrogen as shown below via the ideal gas equation:
Next, one identifies the limiting reagent by computing the moles of hydrogen that completely react with 1.64mol of nitrogen as follows:
In such a way, as there are just 2.65 available moles of hydrogen one states that we have spare nitrogen and the hydrogen is the limiting reagent, thus, the yielded moles of ammonia are computed as:
Finally, one computes the required volume in liters as:
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