Methanol (CH3OH) is the simplest of the alcohols. It is synthesized by the reaction of hydrogen and carbon monoxide
CO(g)+2H2(g)=CH3OH

If 500 mol of CO and 750 mol of H2 are present, which Is the limiting reactant?

By reacting carbon monoxide and hydrogen the formation of methanol takes place, the reaction is,

CO(g)+2H₂(g)⇔CH₃OH (g)

Based on the given reaction, one mole of methanol is obtained by reacting one mole of carbon monoxide (CO) with the two moles of hydrogen (H₂). It is mentioned in the question that for the reaction 500 mol of carbon monoxide and 750 moles of hydrogen are present.

Therefore for 500 moles of carbon monoxide, there is a requirement of 2 × 500 moles of hydrogen, which is equivalent to 1000 moles of hydrogen (H₂). However, only 750 moles of hydrogen is present. Therefore, the limiting reactant in the given case is H₂. The present moles of H₂ will react with 0.5 × 750 moles of CO = 375 mole of CO

The additional or excess concentration of CO, which is the excess reactant will be, 500-375 = 125 moles.

the reaction mixture contains some ammonia, plus a lot of unreacted nitrogen and hydrogen. the mixture is cooled and compressed, causing the ammonia gas to condense into a liquid. the liquefied ammonia is separated and removed. the unreacted nitrogen and hydrogen are then recycled back into the reactor .

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer