Question

10. How much methanol (CH₃OH, in grams) can be formed from 49.2 mg of hydrogen? Assume excess CO.

CO(g) + 2H₂(g) → CH₃OH(g)

-Consider the following balanced chemical equation: 2N₂H₄(g) + N₂O₄(g) → 3N₂(g) + 4H₂O(g)

a. How many moles of water form from 0.438 moles of N₂O₄?

b. How many moles of water form from 0.438 moles of N₂H₄?

c. How many moles of nitrogen form from 0.438 moles of N₂O₄?

d. How many moles of nitrogen form from 0.438 moles of N₂H₄?

Answer 1

10)

Molar mass of H2 = 2.016 g/mol

mass of H2 = 4.92*10^-2 g

mol of H2 = (mass)/(molar mass)

= 4.92*10^-2/2.016

= 2.44*10^-2 mol

Balanced chemical equation is:

CO(g) + 2H2(g) → CH3OH(g)

According to balanced equation

mol of CH3OH formed = (1/2)* moles of H2

= (1/2)*2.44*10^-2

= 1.22*10^-2 mol

Molar mass of CH3OH,

MM = 1*MM(C) + 4*MM(H) + 1*MM(O)

= 1*12.01 + 4*1.008 + 1*16.0

= 32.042 g/mol

mass of CH3OH = number of mol * molar mass

= 1.22*10^-2*32.04

= 0.391 g

Answer: 0.391 g

2)

a)

from given reaction,

moles of H2O formed = (4/1)*moles of N2O4

= 4*0.438 moles

= 1.75 moles

Answer: 1.75 moles

b)

from given reaction,

moles of H2O formed = (4/2)*moles of N2H4

= (4/2)*0.438 moles

= 0.876 moles

Answer: 0.876 moles

C)

from given reaction,

moles of N2 formed = (3/1)*moles of N2O4

= 3*0.438 moles

= 1.31 moles

Answer: 1.31 moles

D)

from given reaction,

moles of H2O formed = (3/2)*moles of N2H4

= (3/2)*0.438 moles

= 0.657 moles

Answer: 0.657 moles