Question

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?

Answer 1

`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.