A) what is the mass in grams of a sample of manganese (ii) phosphite containing 7.23 x 10^22 phosphite ions?

b) determine the percent composition (by mass) for each element in ammonium dichromate (nh4)2cr2o7

c) a compound with a molecular mass of 202.23g/mol was found to have the following mass percent composition: 95.02% carbon and 4.98% hydrogen. determine its:

*empirical formula

*molecular formula

Explanation:

a) Manganese (II) phosphite :

Number of phosphite ions in manganese (II) phosphite =

In one molecule of manganese (II) phosphite there are 2 ions of phosphite.

Then ions of phosphite will be ions will be in:

Moles of manganese (II) phosphite;=

Mass of 0.2401 mol of Manganese (II) phosphite :

0.2401 mol × 322.75 g/mol = 77.49 g

77.49 is the mass in grams of a sample of manganese (II) phosphite.

b) Molar mass of ammonium chromate =252.07 g/mol

Percentage of Nitrogen:

Percentage of hydrogen:

Percentage of chromium:

Percentage of oxygen:

c) Molar mass of the substance = 202.23 g/mol

Percentage of the hydrogen = 4.98 %

Let the molecular formula be

Percentage of the carbon = 95.02 %

x= 16.01 ≈ 16

Percentage of the hydrogen= 4.98 %

y= 10.07 ≈ 10

The molecular formula of the substance is .

The empirical formula of the substance is .

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