Answer:
19. grams is the minimum mass of butane that could be left over by the chemical reaction.
Explanation:
[tex]2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O[/tex]
Moles of butane = [tex]\frac{35.5 g}{58 g/mol}=0.61 mol[/tex]
Moles of oxygen = [tex]\frac{33.}{18 g/mol}=1.8 mol[/tex]
According to reaction, 13 mole of oxygen reacts with 2 moles of butane ,then 1.8 moles of oxygen will react with :
[tex]\frac{2}{13}\times 1.8mol=0.28 mol[/tex] of butane
Moles of butane left unreacted = 0.61 mol - 0.28 mol = 0.33 mol
Mass of 0.3301 moles of butane :
0.33 mol × 58 g/mol = 19. g
19. grams is the minimum mass of butane that could be left over by the chemical reaction.
