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a special engine can burn hydrogen gas in air to produce water vapor. suppose the fuel tank contains 150.0 l of h2 gas at 20.0 atm pressure at 25oc. if all of this gas is burned, what mass of water vapor is produced? h2 o2 --> h2o round your answer to two decimals.

User Ziur Olpa
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The balanced chemical equation for the combustion of hydrogen gas is:

2H2(g) + O2(g) → 2H2O(g)

From the equation, we see that 2 moles of water are produced for every 2 moles of hydrogen gas burned. Therefore, the number of moles of water produced can be calculated as:

moles of H2O = moles of H2/2

To find the number of moles of hydrogen gas, we can use the ideal gas law:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. Rearranging this equation, we get:

n = PV/RT

Plugging in the given values, we get:

n = (20.0 atm * 150.0 L) / (0.0821 L·atm/(mol·K) * 298 K) = 123.1 mol

Therefore, the number of moles of water produced is:

moles of H2O = 123.1 mol / 2 = 61.6 mol

Finally, we can convert the number of moles of water to mass using the molar mass of water:

mass of H2O = moles of H2O * molar mass of H2O

mass of H2O = 61.6 mol * 18.015 g/mol = 1109.98 g

Rounding to two decimal places, the mass of water vapor produced is 1110.00 g.
User Lazar
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