Decaborane, B 10 H 14 , was used as a fuel forrockets in the 1950\'s . It ignite
ID: 690984 • Letter: D
Question
Decaborane, B10H14, was used as a fuel forrockets in the 1950's . It ignites spontaneously with O2according to the following reaction: B10H14 +11 O2 -> 5B2O3 + 7H2O If a rocket's fuel tanks contain9.636×105g of decaborane and1.800×105g of oxygen,* which of the reactants isthe limiting reagent?1.) Oxygen
2.) Decaborane
3.) Neither
**What mass of the other reactant is left over after the limitingreagent in the above mixture has been completely consumed?
***The best design for rockets is one in which both fuel materialsrun out at the same time to minimize excess mass. If the total massof both components in a decaborane/oxygen rocket must be4.291×106g, what mass of liquid oxygen should beused?
Explanation / Answer
First, you need to find the molar mass of both compounds by addingup the atomic mass (look at the periodic table) of each of theirelements: B10H14 = (10.811)*10 + (1.008)*14 = (108.11 +14.112) = 122.22 g/mol O2 = (15.999)*2 = 32.000 Now, convert both of the gram measurements to moles: (9.636×105g decaborane) / (122.222 g/mol) = 7884.0moles of decaborane (1.800×105g O2) / (32.000 g/mol) = 5625.0 moles ofoxygen Now, look at the ratio of decaborane to oxygen in the originalformula. 11 moles of O2 is burned for every 1 mole ofdecaborane. Would there be enough oxygen here to satisfy thedecaborane? Is there 11 times more oxygen thandecaborane? If not, that would mean there is not enoughoxygen, right? If this reaction proceeds, there will still besome decaborane left. The reaction is limited by the compoundthat 'runs out' first. I hope that gets you started. For the other part of thequestion (**): You will need to use the ratio of O2 to decaborane again. Only 1/11 of the oxygen moles will be used for every 1 mole ofdecaborane. That means that (5625.0 / 11) moles of decaboranewill be used. For (***; the third part): You can make an algebraic equation out of this if you wish. The total weight will be the weight of the O2 and the DB together(just add them). Now, we just need equations for the mass'sof each compound. The number of moles of oxygen required is11 times that of DB. See how many grams of each are in a 1:1 ratio. That's easy,right? (122.222 g of DB and 32.000*11=352 g of O2) Now, make that into an equation: 4.291 x 106 g = (122.222g DB + 352g O2) *Factor? In other words, what multiplication factor of this ratio can we useto reach the big number? Well, just divide both sides by(122.222 + 352): 9048.50 = factor Now, just multiply the 352 g of O2 by that factor: (9048.50 * 352) = 3.185 x 106 g of O2 is needed. To check this answer just make sure that # of moles of O2 (3,185 x106 g O2 / 32.000 g/mol) is 11 times as much as thenumber of moles of DB (4.291 x 106 - 3.185 x106)g DB / 122.222 mol/g). Note that because ofrounding it might be a little more or a little less than 11times. There is probably an easier way to figure the last part of thisproblem but I can't think of it right now. Good luck!
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