How many grams of p2o3 can be produced from 95l o2 at STP

Explanation:

Balanced Equation

P

2

O

3

+

3H

2

O

→

2H

3

PO

3

First use the balanced equation to determine the mole ratio between

H

3

PO

3

and

P

2

O

3

. This ratio will be used to determine the moles

P

2

O

3

required to produce

10.2

moles

H

3

PO

3

.

Mole Ratio Between

P

2

O

3

and

H

3

PO

3

from the balanced equation.

1

mol P

2

O

3

2

mol H

3

PO

3

and

2

mol H

3

PO

3

1

mol P

2

O

3

Multiply the moles

H

3

PO

3

by the molar mass that cancels

H

3

PO

3

and leaves

P

2

O

3

10.2

mol H

3

PO

3

×

1

mol P

2

O

3

2

mol H

3

PO

3

=

5.10 mol P

2

O

3

Now that the moles

P

2

O

3

required to produce

10.2 mol H

3

PO

3

are known, multiply the number of moles by its molar mass,

109.945 g/mol

. https://www.ncbi.nlm.nih.gov/pccompound?term=P2O3 This will give the mass in grams needed for

P

2

O

3

to produce

10.2 mol H

3

PO

3

.

5.10

mol P

2

O

3

×

109.945

g P

2

O

3

1

mol P

2

O

3

=

561 g P

2

O

3

rounded to three significant figures

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

i believe the answer is true.