When propane, $C_{3} H_{8}$ $C_3H_8$ , is burned, carbon dioxide and water vapor are produced according to the following reaction: $C_{3} H_{8} + 5 O_{2} \to 3 C O_{2} + 4 H_{2} O$ $C_3H_8 + 5O_2 -> 3CO_2 + 4H_2O$ . 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?

Question

When propane, $C_{3} H_{8}$ $C_3H_8$ , is burned, carbon dioxide and water vapor are produced according to the following reaction: $C_{3} H_{8} + 5 O_{2} \to 3 C O_{2} + 4 H_{2} O$ $C_3H_8 + 5O_2 -> 3CO_2 + 4H_2O$ . 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?

1 Answer

Impact of this question

8703 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-19T08:29:53

$C_{3} H_{8} + 5 O_{2} \to 3 C O_{2} + 4 H_{2} O$ $C_3H_8+5O_2 rarr 3CO_2 + 4H_2O$

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

Explanation:

$Moles of carbon dioxide:$ $"Moles of carbon dioxide:"$ $\frac{132 \cdot g}{44.0 \cdot g \cdot m o l^{- 1}} = 3 \cdot m o l$ $(132*g)/(44.0*g*mol^-1)=3*mol$

$Moles of water:$ $"Moles of water:"$ $\frac{72 \cdot g}{18.01 \cdot g \cdot m o l^{- 1}} = 4 \cdot m o l$ $(72*g)/(18.01*g*mol^-1)=4*mol$

$Moles of oxygen:$ $"Moles of oxygen:"$ $\frac{160 \cdot g}{32.0 \cdot g \cdot m o l^{- 1}} = 5 \cdot m o l$ $(160*g)/(32.0*g*mol^-1)=5*mol$

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

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

$C_{5} H_{12} + 8 O_{2} \to 5 C O_{2} + 6 H_{2} O$ $C_5H_12 + 8O_2 rarr 5CO_2 + 6H_2O$

Science

Math

Humanities

... and beyond

1 Answer

Explanation:

Related questions

Impact of this question