Producing H_{2} Gas: Na + H_{2}O Reaction

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SUMMARY

The reaction between sodium (Na) and water (H2O) produces hydrogen gas (H2) according to the equation 2Na + 2H2O → 2NaOH + H2. For 1.74 moles of Na, the correct calculation involves stoichiometric coefficients, which indicate that 1 mole of Na produces 0.5 moles of H2. Therefore, 1.74 moles of Na will yield 0.87 moles of H2. At 280 K and 96 kPa, the volume of hydrogen gas produced can be calculated using the ideal gas law.

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1. How many liters of hydrogen gas will be produced at 280 K and 96 kPa if 1.74 of Na reacts with excess water according to the following equation.

2Na + 2H[itex]_{2}[/itex]O [itex]\rightarrow[/itex]2NaOH + H[itex]_{2}[/itex]


Im gussing I am going to have to find the number of moles of H[itex]_{2}[/itex] before I can do anything.

So I am thinking that since there is 1.74 mol of Na that is going to be the number of moles of H[itex]_{2}[/itex]. Is that right?
 
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Use [noparse] and [/noparse] tags to format subscripts and superscripts (so [noparse]H3O+[/noparse] yields H3O+) - don't use inline LaTeX for formatting single characters.

You are wrong, but not far from the correct answer. You forgot to take stoichiometric coefficients into account. Please read about stoichiometric calculations.
 

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