When propane, $C_3H_8$ , is burned, carbon dioxide and water vapor are produced according to the following reaction: $C_3H_8 + 5O_2 -> 3CO_2 + 4H_2O$ . How much propane is burned if 160.0 g of O2 are used and 132 g of CO2 and 72.0 g of H2O are produced?

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Answer 1 · 2016-12-19T08:29:53

$C_3H_8+5O_2 rarr 3CO_2 + 4H_2O$

Precisely one mole, $44*g$ of pentane, were combusted.

$"Moles of carbon dioxide:"$ $(132*g)/(44.0*g*mol^-1)=3*mol$

$"Moles of water:"$ $(72*g)/(18.01*g*mol^-1)=4*mol$

$"Moles of oxygen:"$ $(160*g)/(32.0*g*mol^-1)=5*mol$

Because there were $3$ $mol$ carbon dioxide produced, and stoichiometric water, one mole precisely of propane was combusted.

If pentane were burned in excess dioxygen, and $220*g$ carbon dioxide were produced, what was the starting mass of pentane?

$C_5H_12 + 8O_2 rarr 5CO_2 + 6H_2O$

When propane, C_3H_8C_3H_8, is burned, carbon dioxide and water vapor are produced according to the following reaction: C_3H_8 + 5O_2 -> 3CO_2 + 4H_2OC_3H_8 + 5O_2 -> 3CO_2 + 4H_2O. How much propane is burned if 160.0 g of O2 are used and 132 g of CO2 and 72.0 g of H2O are produced?