Answer to Question #106391 in General Chemistry for Logan Benton

Question

In the reaction CO(g)+ 2H2(g)-> CH3OH(l)
a. Which reactant is the limiting reactant if 356g of CO and 65.0g of H2 are mixed and allowed to react?
b. How many grams of methanol can be produced from the conditions in part a?

Answer 1

1. In the reaction

 $CO+2H_2=CH_3 OH$

a. Which reactant is the limiting reactant if 356g of CO and 65.0g of H2 are mixed and allowed to react?

b. How many grams of methanol can be produced from the conditions in part a?

Solution:

Step 1 Balance the equation

$CO+2H_2=CH_3 OH$

Step 2 Apply mole concept

$1mole CO+2 mole H_2=1 mole CH_3 OH$

 $W_m=$ molecular weight

$28g CO+4g H_2=32g CH_3 OH$

hence

$W_m (CO):W_m (H_2)=28:4=7:1$

Step 3  Compare with given amount of reagents

$W(CO)=356gm$

$W(H_2)=65g m$

 $W(CO): W(H_2)=356:65 =5.47:1$

$W(CO): W(H_2) < W_m (CO):W_m (H_2)$

 

Answer : Hence CO is limiting reagent

b)

1mole CO produce 1 mole methanol

28g CO produce 32g methanol

356g CO produce  32/28×356 g methanol=406.86g methanol

Answer : 406.86 g Methanol will be produced