2H2+O2=H=-121k?1molH2O
how many grams of hydrogen are needed to produce 1.20x10power3 Kj of heat?
121 kJ/mol x xgrams/4 = 1.20E3 kJ.
Solve for x.
To answer this question, we first need to calculate the number of moles of water produced in the given reaction.
From the balanced chemical equation:
2H2 + O2 → 2H2O
We can see that 2 moles of hydrogen gas (H2) are required to produce 2 moles of water (H2O).
Now, we can use the given enthalpy change of the reaction (-121 kJ/mol H2O) to calculate the amount of heat released when 1 mole of water is formed.
1 mole H2O → -121 kJ
To calculate the amount of heat released when 1.20 x 10^3 kJ is produced, we can set up a proportion:
1 mole H2O / -121 kJ = x moles H2O / 1.20 x 10^3 kJ
Cross-multiplying, we get:
1.20 x 10^3 kJ x 1 mole H2O = -121 kJ x moles H2O
Rearranging the equation, we have:
moles H2O = (1.20 x 10^3 kJ / -121 kJ)
moles H2O = -9.92
Since we cannot have a negative number of moles, it seems there was a mistake in the initial question or calculation.
However, if the question was meant to ask how many grams of hydrogen are needed to produce 1.20 x 10^3 kJ of heat, we can use the molar mass of hydrogen (H2) to determine the grams required.
The molar mass of H2 is approximately 2 g/mol.
To calculate the grams of hydrogen required, we can use the mole-to-gram conversion:
moles H2 x molar mass H2 = grams H2
Using the calculated number of moles from before (moles H2 = -9.92), we have:
-9.92 mol H2 x 2 g/mol = -19.84 g H2
Again, we encounter a negative value, which suggests either a mistake in the initial question or calculation.