The complete combustion of heptane (C7H16) in oxygen produces carbon dioxide and water. If I had 4.75g of heptane and combusted this to completion, how many grams of carbon dioxide would I produce?

Grams of water produced=3.72g

2C7H16 + 22O2 => 14CO2 + 16H2O

CH3(CH2)5CH3CH3CH(CH3)(CH2)3CH33 methyl haxane2,2 di methyl pentane2,4 dimethyl pentane3,3 dimethyl pentane2,2,3 dimethyl trimethyl butane3, ethyl pentane2,3 dimethyl pentane

You can calculate this by such reaction: NH₄NO₃ --gt; N₂ + frac12;O₂ + 2H₂O ==gt; 2NH₄NO₃ --gt; 2N₂ + O₂ + 4H₂O The law of conservation of mass solves this...

m(SO2) = 29 g M(SO2) = 32 + 16*2 = 64 g/mol n(SO2) = m/M = 29g/64g/mol = 0.453 mol According to the following reaction 1 mol of sulfur dioxide...

5n(C5H12)=n(CO2)m(CO2)=44,00 g/mole * n(CO2)m(C5H12)=72,00 g/mole * n(C5H12)n(C5H12)=4,444 molesn(CO2)=22,22 molesm(CO2)=977,68 g

CH4+2O2=CO2+2H2On(CO2):n(CH4)=1:1=2.5molm(CO2)=n(CO2)*M(CO2)=2.5*44=110gȠ=(mpr/mt)*100%=(77/110)/100%=70%Answer: Ƞ=70%

C+O2=CO23935 x 60 /12 = 19 675 j/mol = 19,675 kj/mol.

CH4(g) + 2 O2 (g) = CO2(g) + 2 H2O (g)1 : 2 : 1 : 223g : 42g : ? : 37g Therefore : ( 23/42) =0.5 moles 1...

2C4 H10 + 13 O2 = 8 CO2 +10 H2On (C4 H10 )= 0,67 mole n (H2O)=10*0.67/2=3,35 moleanswer: 3,35 mole