1.- Nitric oxide (NO) reacts with oxygen gas to form nitrogendioxide (NO 2 ), a

Question

1.- Nitric oxide (NO) reacts with oxygen gas to form nitrogendioxide (NO2), a dark-brown gas: 2NO(g) + O2(g) -----> 2NO2(g) In one experiment 0.886 mole of NO is mixed with 0.503 mole ofO2. Determine which of the two reactants is the limitingreactant. Calculate also the number of moles of NO2produced. 2.- Nitroglycerin(C3H5N3O9) is apowerful explosive. Its descomposition may be represented by 4C3H5N3O9------> 6N2 + 12CO2 +10H2O + O2    This reactiongenerates a large amount of heat and gaseous products. It is thesudden formation of these gases , together with their rapidexpansion, that produces the explosion. a) What is the maximum amount of O2 in grams thatcan be obtained from 2.00 x 10^2 g of nitroglycerin? b)Calculate the % yield in thsi reaction if the amount ofO2 generated is found to be 6.55 g.   1.- Nitric oxide (NO) reacts with oxygen gas to form nitrogendioxide (NO2), a dark-brown gas: 2NO(g) + O2(g) -----> 2NO2(g) In one experiment 0.886 mole of NO is mixed with 0.503 mole ofO2. Determine which of the two reactants is the limitingreactant. Calculate also the number of moles of NO2produced. 2.- Nitroglycerin(C3H5N3O9) is apowerful explosive. Its descomposition may be represented by 4C3H5N3O9------> 6N2 + 12CO2 +10H2O + O2    This reactiongenerates a large amount of heat and gaseous products. It is thesudden formation of these gases , together with their rapidexpansion, that produces the explosion. a) What is the maximum amount of O2 in grams thatcan be obtained from 2.00 x 10^2 g of nitroglycerin? b)Calculate the % yield in thsi reaction if the amount ofO2 generated is found to be 6.55 g.  

Explanation / Answer

In each case, the limiting reagent is the one with thelowest number of moles. (moles = 'n') 1. 0.886 mol NO > 0.503 mol O2, so oxygen gasis limiting. 2 moles of NO2 gas are produced for everymole of O2, so: n(NO2) = 2*n(O2) = 2*0.503 = 1.006 moles 2. n(nitroglycerine,C3H5N3O9) = mass /molar mass = 200 g / (12*3+1*5+14*3+16*9) = 200 g / 227 gmol-1 = 0.881 moles (a) Note that 1 mole of O2 is produced for every 4 molesof nitroglycerine, so: n(O2) = n(nitroglycerine) / 4 = 0.881 / 4 = 0.220moles mass(O2) = moles * molar mass = 0.220 mol * 32 gmol-1 = 7.05 g (b) % Yield = Actual Yield / Theoretical Yield * 100% = 6.55 / 7.05* 100 = 92.9%

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