What mass of methanol (CH3OH) is produced when 140.1 g of carbon monoxide reacts with 12.12 g of hydrogen?
co(g)+2h2(g)--->ch3oh(/)
a96.12g
b160.2g
c192.2g
d6.06.g

Explanation:This is a limiting reactant problem.  One way to deal with limiting reactant problems is to compute the amount of product using each of the reactants.  The theoretical yield will be the lesser of the two and that will also indicate the limiting reactant.

CO(g) + 2H2(g) --catalysts--> CH3OH(g)

140.1g..12.12g ?g

140.1g CO x (1 mol CO / 28.0g CO) x (1 mol CH3OH / 1 mol CO) x (32.0g CH3OH / 1 mol CH3OH) = 160.1g CH3OH

12.12g H2 x (1 mol H2 / 2.02g H2) x (1 mol CH3OH / 2 mol H2) x (32.0g CH3OH / 1 mol CH3OH) = 96.00g CH3OH

96.00g CH3OH will be produced and hydrogen is the limiting reactant.

c is the answer because when something is left after something evaporates that shows the chemical change.

while cooking a piece of 1/2 pound beef steak, a cast iron pan (2.4kg in weight) was heated 260 celsius.