Try it on paper first. A hint, the key idea and the worked solution are here when you need them.
Ethyne is prepared in a flask by heating of 1,2-dibromoethane with an excess of alcoholic potassium hydroxide. The reaction is . Use for 1,2-dibromoethane, for KOH and for KBr. One mole of gas occupies at standard temperature and pressure. Find the mass of KOH that reacts, the volume of ethyne at standard temperature and pressure, and the mass of KBr formed, if the reaction is complete.
Elimination takes a beta hydrogen and the halogen away and leaves a double bond. The more substituted alkene is the main product (Saytzeff). Each double bond with different groups on both ends counts as cis and trans. Alcoholic KOH and high temperature favour elimination. The ratio of two rates is .
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Step 1. Amount of dibromide. .
Step 2. KOH. Two moles per mole: .
Step 3. Ethyne. One mole per mole: .
Step 4. KBr. Two moles per mole: .
Answer: of KOH, of ethyne and of KBr.
Check. Mass balance: of reactants. The products are of ethyne, of KBr and of water. That is .
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