Hydrogen and oxygen react chemically to form water. How much water would form if 4.8 grams of hydrogen reacted with 38.4 grams of oxygen?

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Answer 1 · 2016-09-05T06:24:12

Approx. $36*g$

We need a stoichiometric equation for water synthesis:

$H_2(g) + 1/2O_2(g) rarr H_2O(l)$

Clearly, dihydrogen must be present in a 2:1 molar ratio with respect to dioxygen.

$"Moles of dihydrogen"$ $=$ $(4.8*g)/(2.01*g*mol^-1)$ $=$ $2.39*mol$

$"Moles of dioxygen"$ $=$ $(38.4*g)/(32.0*g*mol^-1)$ $=$ $1.2*mol$

Given these molar quantities (they are approx. 2:1) the gases are present in stoichiometric proportion. $2$ $mol$ water are going to be produced.

$2*molxx18.00*g*mol^-1$ $=$ $36*g$

Do you think that this would be an exothermic reaction; i.e. do you think that energy would be released by the reaction?