How do you calculate the mass of water produced from the reaction of 24.0 g of #H_2# and 160.0 g of #CO_2#? What is the limiting reagent?

1 Answer
anor277
Mar 14, 2017

#CO_2(g) + 4H_2(g) rarr CH_4(g) + 2H_2O(g) + Delta#
#DeltaH = −165.0 *kJ*mol^-1#

Explanation:

I assume this is the reaction #"(the Sabatier reaction)"# to which you refer.

#"Moles of dihydrogen"=(24.0*g)/(2.02*g*mol^-1)=11.9*mol#

#"Moles of carbon dioxide"=(160.0*g)/(44.0*g*mol^-1)=3.64*mol#

Clearly, #H_2# is the reagent in deficiency, and according to this stoichiometry, #2.98*mol# carbon dioxide will react. And given this equivalence, #2xx2.98*mol# water will result, i.e. this represents a mass of #2xx2.98*molxx18.0*g*mol^-1=107.1*g#.

This reaction requires substantial catalysis, usually in the form of a ruthenium complex.

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