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kati45 [8]
3 years ago
6

N—butane fuel (c4h10) is burned with a stoichiometric amount of air. determme the mass fraction of each product. also, calculate

the mass of carbon dioxide in the products and mass of air required when 6.14 kg of fuel is burned.
Chemistry
1 answer:
harina [27]3 years ago
8 0
The balanced equation for the burning of N-butane is;
                      
                          2C₄H₁₀  +  13O₂  →   8CO₂   +  10H₂O
stoi. ratio               2    :         13       :      8        :      10
molar mass          58              32               44           18
moles                    X             13 X/2        8X/2          10X/2
                            105.86      688.10      423.45       529.31
produced mass                                    18632.8 g    9527.6 g

moles of n-butane = X = 6.14 x 10³ g / 58 g mol⁻¹ 
hence other moles of compounds can be calculated.

Mass = molar mass * moles

mass fraction = mass / total mass

hence the mass fraction of
    CO₂ = 18632.8 / 6.14 x 10³  = 3.03
    H₂O = 9527.6 / 6.14 x 10³    = 1.55

the required mass of air = 688.10 x 32 = 22019.2 g = 2.20 kg



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As in the given case, heat is released by the system, so sign of q is negative, or q < 0

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In the given case, work is done on the system, therefore sign of w is positive, or w > 0

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