In the equation #2C_2H_2(5) + 5O_2(5) -> 4CO_2(g) + 2H_2O(g)#, how many grams of oxygen gas are needed for the complete combustion of #1.8 * 10^2# #g# of acetylene?

1 Answer
Jan 6, 2016

Answer:

About #553.6# grams of oxygen are used in this combustion.

Explanation:

Molecular mass of acetylene: #2 * MM_C + 2 * MM_H#
#MM_C# = Molecular mass of carbon
#MM_H# = Molecular mass of hydrogen

So, #2 * 12 + 2 * 1 = 26# grams/mole

#"Mass used in combustion"/"molecular mass" = "moles used in combustion"#

#(180cancel(" grams"))/(26 cancel(" grams")/"mole")= 6.92 " moles"#

According to the main equation, to 2 moles of acetylene, 5 moles of oxygen are used. #MM_O = 2 * 16 = 32# grams/mole

So, if we use #6.92# moles of Acetylene, #x# moles of oxygen will be necessary.

#x = (6.92 * 5)/2 = 3.46 * 5 = 17.3# moles of oxygen.
#17.3 cancel("moles") * 32 "grams"/cancel("mole") = 553.6# grams of oxygen.