g Acetylene gas is often used in welding torches because of the very high heat produced when it reacts with oxygen gas, producing carbon dioxide gas and water vapor. Calculate the moles of carbon dioxide produced by the reaction of of acetylene. Be sure your answer has a unit symbol, if necessary, and round it to the correct number of significant digits.

3.0 mol CO₂

Explanation:

Acetylene gas is often used in welding torches because of the very high heat produced when it reacts with oxygen gas, producing carbon dioxide gas and water vapor. Calculate the moles of carbon dioxide produced by the reaction of 1.5 moles of acetylene. Be sure your answer has a unit symbol, if necessary, and round it to the correct number of significant digits.

Step 1: Write the balanced combustion reaction

C₂H₂(g) + 2.5 O₂(g) ⇒ 2 CO₂(g) + H₂O(g)

Step 2: Establish the appropriate molar ratio

According to the balanced equation, the molar ratio of C₂H₂ to CO₂ is 1:2.

Step 3: Calculate the moles of CO₂ produced from 1.5 moles of C₂H₂

1.5 mol C₂H₂ × 2 mol CO₂/1 mol C₂H₂ = 3.0 mol CO₂

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.