NCMom Posted December 11, 2014 Share Posted December 11, 2014 I'm not a big fan of teaching chemistry from random worksheets but boy this teacher sure is. All of the other problems were pretty easy to check but student and I are getting different answers on this one. I got 2 hours of sleep so that may be the real problem lol. Barium reacts with water. If 25g of barium and 25 g of water react, what volume of hydrogen gas is produced? Thank in advance. Georgia Quote Link to comment Share on other sites More sharing options...
Twigs Posted December 12, 2014 Share Posted December 12, 2014 … Barium reacts with water. If 25g of barium and 25 g of water react, what volume of hydrogen gas is produced? … This is a limiting reactant problem. 1st, you need the balanced chemical equation: Ba + 2 H2O → Ba(OH)2 + H2 To figure out the mass of H2 produced, calculate how much H2 is produced from each mass of reactants. The general calculation is: (mass of reactant) x (1/molar mass of reactant) x (mole-to-mole ratio) x (molar mass of product) For Ba: (25 g Ba) x (1 mol Ba/137.3 g Ba) x (1 mol H2/1 mol Ba) x (2.016 g H2/1 mol H2) = 0.37 g H2 For H2O: (25 g H2O) x (1 mol H2O/18.02 g H2O) x (1 mol H2/2 mol H2O) x (2.016 g H2/1 mol H2) = 1.4 g H2 25 g Ba will yield 0.37 g H2 25 g H2O will yield 1.4 g H2 Since the reaction stops when the 25 g Barium are used up, the Barium is the limiting reactant, and 0.37 g of hydrogen gas are produced. Quote Link to comment Share on other sites More sharing options...
Dicentra Posted December 12, 2014 Share Posted December 12, 2014 Twigs beat me to it - I was just starting to type up a response. :) One small correction - I think the original question asked for volume of hydrogen gas and not mass. Assuming they want the volume at STP, you'd take the moles of hydrogen gas produced and multiply by 22.4 L/mol. 0.182 mol x 22.4 L/mol = 4.08 L of hydrogen gas at STP If they want the volume at any other temperature and pressure, they'd have to give you those conditions and then you could use the ideal gas law to calculate volume. :) Quote Link to comment Share on other sites More sharing options...
Twigs Posted December 12, 2014 Share Posted December 12, 2014 Twigs beat me to it - I was just starting to type up a response. :) One small correction - I think the original question asked for volume of hydrogen gas and not mass. Assuming they want the volume at STP, you'd take the moles of hydrogen gas produced and multiply by 22.4 L/mol. 0.182 mol x 22.4 L/mol = 4.08 L of hydrogen gas at STP If they want the volume at any other temperature and pressure, they'd have to give you those conditions and then you could use the ideal gas law to calculate volume. :) :blushing: oops! Thanks, Dicentra, for catching that. Quote Link to comment Share on other sites More sharing options...
Dicentra Posted December 12, 2014 Share Posted December 12, 2014 :blushing: oops! Thanks, Dicentra, for catching that. :D No worries! Your use of sig figs is better than mine - my volume may not be exactly what the OP has in the answer key. I dislike having to think about sig figs. I'm too lazy. ;) Quote Link to comment Share on other sites More sharing options...
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