In: Chemistry
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
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