Consider the unbalanced equation for the combustion of hexane: C_6H_14(g) + O_2(
ID: 940543 • Letter: C
Question
Consider the unbalanced equation for the combustion of hexane: C_6H_14(g) + O_2(g) rightarrow CO_2(g) + H_2O(g) Balance the equation and determine how many moles of O_2 are required to react completely with 7.2 moles of C_6H_14. Consider the unbalanced equation for the neutralization of acetic acid: Balance the equation and determine how many moles of Ba(OH)_2 are required to completely neutralize 0.461 mole of HC_2H_3O_2. Calculate how many moles of NO_2 form when each quantity of reactant completely reacts. Calculate how many moles of NH_3 form when each quantity reactant completely reacts.Explanation / Answer
Question 1)
Balanced equation:
2 C6H14 + 19 O2 12 CO2 + 14 H2O
Reaction type: combustion reaction
Molar mass:
C6H14: 6X12.01+14X1.008= 86.17 g/mol
O2: 32.00 g/mol
7.2 mol C6H14 X 19/2 = 68.4 mol O2
68.4 mol O2 require to react completely with 7.2 mol C6H14
Question 2)
Balanced equation:
2 Ch3COOH + Ba(OH)2 = Ba(Ch3COO)2 + 2 H2O
Reaction type: double replacement
0.461 mol Ch3COOH X 1/2 = 0.2305 mol Ba(OH)2
0.2305 mol Ba(OH)2 require to react completely with 0.461 mol Ch3COOH
Question 3)
2 N2O5(g) = 4 NO2(g) + O2(g)
N2O5 = 108g / mol, NO2= 46g / mol Your basic formula for this is:
Note that for 2 mol N2O5, we are required for 4 mol NO2
moles of NO2 form = 2.5 mol N2O5x[4 mol NO2/2 mol]
moles of NO2 form = 5 mol NO2
b) 6.8 mol N2O5
moles of NO2 form = 6.8 mol N2O5x[4 mol NO2/2 mol ]
moles of NO2 form=13.6 mol NO2
c) 15.2 gm N2O5
Have to find the number of moles N2O5 available:
molar mass N2O5 = 14.01 x 2 + 16.00 x 5 = 108.02 g/mol
so mol N2O5 available = 15.2 g N2O5 x [1 mol / 108.02 g] = 0.1407 mol N2O5
moles of NO2 form = 0.1407 mol N2O5x[4 mol NO2/2 mol]
moles of NO2 form=0.281 mol NO2
d) 2. 87 kg N2O5
number of mol N2O5 available = 2.87 kg N2O5 x [1000 g / kg] x [1 mol / 108.02 g] = 26.56 mol N2O5
moles of NO2 form = 26.26 mol N2O5x[4 mol NO2/2 mol ]
moles of NO2 form=53.12 mol NO2
Question 4)
Balanced equation:
3 N2H4 = 4 NH3 + N2
Reaction type: decomposition
so we get that 1 mole of N2H4 (left) gives (4/3) moles of NH3 on complete reaction.
moles of NH3 form = 2.6 mol N2H4 x [4 mol NH3/3 mol]
moles of NH3 form = 3.46 mol NH3
moles of NH3 form = 3.55 mol N2H4 x [4 mol NH3/3 mol ]
moles of NH3 form = 4.73 mol NH3
Molar mass:
N2H4 : 2X14.01+2 X1.008= 30.016 g/mol
NH3: 17.00 g/mol
so mol N2H4 available = 63.5 g N2H4 x [1 mol / 30.016 g] = 2.11 mol 30.016
moles of NH3 form = 2.11 mol N2H4 x [4 mol NH3/3 mol]
moles of NH3 form= 2.82 mol NH3
number of mol N2H4 available = 4.88 kg N2H4 x [1000 g / kg] x [1 mol / 30.016 g] = 162.57 mol N2H4
moles of NH3 form = 162.57 mol N2H4 x [4 mol NH3/3 mol]
moles of NH3 form= 216.76 mol NH3