Question

What mass of water (in grams) forms from the reaction of 2.22 L of hydrogen gas and 2.24 L of oxygen gas? Both gases are at 726 torr and 7.8 ^oC.

Answer 1

2 H2 + O2 --> 2 H2O

PV=nRT

n = (PV)/(RT)

Pressure = 726 Torr

Volume = 2.22 L

R = gas constant = 62.3637 (L * Torr) / (mol * K)

Temperature = 7.8 C + 273= 280.8K

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n = (PV)/(RT)

n = ( (726 Torr) * (2.22 L) ) / ( 62.3637 (L * Torr)/(mol*K) * 280.8K )

n= 726*2.22/(62.363*280.8)

n= 0.09203 mol H2

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Now do the same with O2 gas

n = ( (726Torr) * (2.24 L) ) / ( 62.3637 (L * Torr)/(mol*K) * 280.8)

n=726*2.24/(62.36*280.8)

n= 0.092865 mol O2
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Now this is just a limiting reactant problem. Use the mole ratios from the balanced equation to calculate how much H2O can be made from each of the moles of reactants. First do H2:

(0.09203 mol H2) * (2 / 2 mol H2O/ mol H2) * (18.0158 g/mol H2O)

= 1.6579 g H2O
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Then O2:

(0.092865 mol O2) * (2 / 1 mol H2O/ mol O2) * (18.0158 g/mol H2O)

= 3.346 g H2O
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Since H2 makes less product, it is the limiting reactant, and so the amount of H2O it makes is the maximum amount that can be produced,

so 1.657 g is your answer.

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