1. Methanol (CH 4 O) can be made by a reaction between carbon monoxide gas and h

Question

1. Methanol (CH4O) can be made by a reaction between carbon monoxide gas and hydrogen gas. Methanol is the only product of the reaction.

a) Starting with 75.0 g of each reactant, calculate the theoretical yield of methanol.

b) How many grams of the excess reactant remains when the reaction is over?

2. Silver(I) chloride is insoluble in water. A chemist wishes to prepare 28.7 g of silver(I) chloride by mixing together aqueous solutions of 0.400 M silver(I) nitrate and 1.00 M calcium chloride. What is the theoretical volume of each solution she should use?

3.What wavelength of light must a hydrogen atom in the ground state absorb for the electron to move to the n = 2 energy level?

please show how you solved the problems

Explanation / Answer

1)

the reaction is given by

CO + 2H2 ---> CH3OH

we know that

moles = mass / molar mass

so

moles of C0 = 75 / 28 = 2.68

moles of H2 = 75 / 2 = 37.5

we can see that

moles of H2 required = 2 x moles of C0

moles of H2 required = 2 x 2.68 = 5.36

but 37.5 moles of H2 is present

so

H2 is in excess and CO is the limiting reagent

now

moles of CH3OH formed = moles of CO reacted = 2.68

mass = moles x molar mass

so

mass of CH3OH = 2.68 x 32

mass of CH3OH = 85.7

so

85.7 grams of methanol is produced

b)

now

excess moles of H2 = 37.5 - 5.36 = 32.14

so

mass of H2 = 32.14 x 2 = 64.28

so

64.28 g of H2 is in excess

2)

the reaction is

CaCl2 + 2 AgN03 ---> Ca(N03)2 + 2 AgCl

moles of AgCl = 28.7 / 143.32 = 0.2

now

moles of AgN03 required = moles of AgCl = 0.2

now

volume (L) = moles / molarity

so

volume of AgN03 = 0.2 / 0.4 = 0.5

so

0.5 L of AgN03 is required

now

moles of CaCl2 = 0.5 x moles of AgN03 = 0.5 x 0.2 = 0.1

so

volume of CaCl2 = 0.1 / 1 = 0.1

so

0.1 L of CaCl2 is required

So finally

500 ml of AgN03 and 100 ml of CaCl2 are required

3)

using rydberg formula for hydrogen

1/lamda = 1.097 x 10^7 [ (1/n1)^2 - (1/n2)^2 ]

in this case

n1 = 1

n2 = 2

so

1/lamda = 1.097 x 10^7 ( 1 - 1/4)

lamda = 1.215 x 10-7

so

the wavelenght is 1.215 x 10-7 m

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