How many moles of oxygen would have be consumed to produce 68.1 g of water in the reaction $C_3H_8 + 5O_2 -> 3CO_2 + 4H_2O$ ?

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Answer 1 · 2016-07-08T15:25:50

Approx. $20*mol" dioxygen gas"$

$"Moles of water"$ $=$ $(68.1*g)/(18.01*g*mol^-1)$ $=$ $3.78*mol$

Given the stoichiometry, clearly there are $5/4xx3.78*mol$ $"dioxygen gas"$ required. What is the equivalent mass?

How many moles of oxygen would have be consumed to produce 68.1 g of water in the reaction C_3H_8 + 5O_2 -> 3CO_2 + 4H_2OC_3H_8 + 5O_2 -> 3CO_2 + 4H_2O?