Mass of watch glass + filter paper = 105.98 g Mass of watch glass + filter paper + crystallized product = 109.03 g
Mass of uncrystallized product (show work) =
Mass of methyl benzoate = 3.08 g
Volume of nitric acid used = 2.0 mL

Theoretical yield based on each of the starting materials

(Please use two Dimensional Analysis (DA) equations, one for the maximum amount of product obtainable from the amount of methyl benzoate you used and the other from the concentrated nitric acid, then use the lesser of the two to determine the Limiting Reagent; you must determine the number of moles in 2.00 mL of concentrated nitric acid [concentration 69.0% (w/w), and density (1.42 g/mL)].

Required:
a. Identity of the Limiting reagent (LR) based on the above two DA equations =
b. Max amount of product obtainable from the LR =
c. Mass of the product you obtained:

Explanation:

Mass of uncrystallized product = (Mass of watch glass + filter paper + crystallized product) - (Mass of watch glass + filter paper)  

Mass of uncreystaliized product = 109.03 gm - 105.98 gm

Mass of uncrystaliized product = 3.05 gm

For methyl benzoate;

mass = 3.08 g

no of moles = 3.08 g/ 136.15 g/mol = 0.0226 mole

It is possible for the formation of 1 mole of nitro methyl benzoate from a mole of methyl benzoate.

moles of nitro methyl benzoate that can be formed from 0.0226 moles of methyl benzoate = 0.0226 moles

∴

mass of nitro methyl benzoate = 0.0226 mol × 181.15 gm /mol

mass of nitro methyl benzoate = 4.098 gm

For HNO_3 solution:

mass = 1.42 gm/ml × 2.0 ml

mass = 2.84 gm

Mass of HNO3 in 2.84 gm solution

= 1.9596 gm

Moles of HNO3 =

= 0.0311 mole

1 mol of HNO_3 can be formed from 1 mole of nitro methyl benzoate

Thus; moles of nitro methyl benzoate that can be formed from 0.0311 mole of HNO_3 = 0.02256 mole

The mass for nitro methyl benzoate can now be determined as:

= 0.0311 mole × 181.5 gm/mole

= 5.634 gm

Since the mass formed from methyl benzoate is lesser, then methyl benzoate serves as the limiting reagent.

The mass obtainable from the LR = 4.098 gm

mark him as !

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g