20- Given the equation
P2O5 + H2O → H3PO4
a)
Balance the equation.
P2O5 + 3 H2O → 2 H3PO4
Phosphorus Pentoxide + Water
→
Phosphoric Acid
Reaction Type: Synthesis
b)
How many moles of H3PO4 would be produced from 108 grams of water?
To solve this Mole-Mass stoichiometry problem we have firstly balance the unbalanced equation. Herein
in this problem we identify that we must find (x) number of moles of H3PO4 that can be produced from
108 grams of water. Secondly, we write the identified elements below the relevant formulas in the
balanced equation as follow:
P2O5 + 3 H2O → 2 H3PO4
108 g
x mol
Thirdly, we find the GFM of H2O
[ 2H+1O= (2x1) + (1x16)=18 g/mol ] to use it in the factor-label
method (dimensional analysis) based on our knowledge of chemistry as follow:
x mol H3PO4 = (108 g H2O) x
(1 mol H2O
(18 g H20)
from GFM
x
(2 mol H3PO4)
( 3 mol H2O)
= 4 mol H3PO4
from the balanced equation
Finally, we cross-out the cross-multiplying units to be left out with the required (final) unit and then do
the mathematical calculations to get the final answer.
Answer= 4 mol H3PO4
c)
How many grams of H3PO4 would be produced from 0.400 moles of P2O5?
To solve this Mass-Mole stoichiometry problem we have firstly balance the unbalanced equation.
Herein, in this problem we identify that we must find (x) number of grams of H3PO4 that can be
produced from 0.400 moles of P2O5. Secondly, we write the identified elements below the relevant
formulas in the balanced equation as follow:
P2O5
+
3 H2O → 2 H3PO4
0.400 mol
x grams
Thirdly, we find the GFM of H3PO4
[ (3H+1P+4 O)= (3x1) + (1x31)+(4x16)=98 g/mol ] to use it
in the factor-label method (dimensional analysis) based on our knowledge of chemistry as
follow:
x g H3PO4 = (0.400 mol P2O5) x
(2 mol H3PO4
(1 mol P2O5)
x
from the balanced equation
(98 g H3PO4)
( 1 mol H3PO4)
= 78.4 g H3PO4
from GFM
Finally, we cross-out the cross-multiplying units to be left out with the required (final) unit and then do
the mathematical calculations to get the final answer.
Answer= 78.4 g H3PO4
d)
How many grams of water would be needed to react with 71.0 grams of P2O5?
To solve this Mass-Mass Stoichiometry problem we must firstly balance if the equation is unbalance; identify
what we are given and what we are asked to answer. Herein we identify that we must find (x) number of grams
of H2O that will be needed to react with 71. 0 grams of P2O5. Secondly, we write the identified elements below
the relevant formulas in the balanced equation as follow:
P2O5
71.0 grams
+
3 H2O → 2 H3PO4
x grams
Thirdly, we find the GFM of P2O5
and
2 P=2x31=62
5 O= 5x16=80__
142g/mol
H2O
as follow:
2 H= 2x1=2
1 O= 1x16=16
18g/mol
Fourthly, we use factor-label method (dimensional analysis) based on our knowledge of chemistry as follow
x g H2O = 71.0 g P2O5 x
(1 mol P2O5)
(142 g P2O5)
x
from GFM
(3 mol H2O)
(1 mol P2O5)
x
from the balanced equation
(18 g H2O)
(1 mol H2O)
= 27 g H2O
from GFM
Finally, we cross-out the cross-multiplying units to be left out with the required (final) unit and then do the
mathematical calculations to get the final answer.
Answer = 27 g H2O
21- If 100 grams of water reacts with 100 grams of calcium carbide according to the following reaction:
CaC2 (s) + 2 H2O (l) → Ca(OH)2 (aq)+ C2H2 (g)
a) Which reactant is the limiting reactant (reagent)?
To solve this problem, firstly, we will find the GFM of both reactants and then convert the given grams
of both reactants into moles.
GFM of CaC2 is [ 1Ca +2C= (1x40)+(2x12)= 64g/mol]
GFM of H2O is [ 2H+1O= (2x1) + (1x16)=18 g/mol ]
mole of CaC2 = 100 g of CaC2 x 1mole of CaC2 = 1.56 moles CaC2
64 g of CaC2
and moles of H2O = 100 g H2O x 1mole H2O = 5.56 moles H2O
18 g H2O
Secondly, we will divide calculated moles of the both reactants by the smallest number (1.56)
1.56 moles CaC2 = 1 moles CaC2
1.56
and
5.56 moles H2O = 3.56 moles H2O
1.56
Thirdly we will write the calculated moles ratio right below the respective reactants in the given
equation.
1 CaC2 (s) + 2 H2O (l) → Ca(OH)2 (aq)+ C2H2 (g)
1 mole
3.56 mole
Per the given balanced chemical equation 1 mole of CaC2 (s) reacts with 2 moles of H2O (l). However, the
calculated ratio of moles clearly shows that for the 1 mole of CaC2 there are 3.56 moles of water
available, which are way more than what the reaction needs. Therefore, it is obvious that the CaC2 is the
limiting reactant because after its consumption the chemical reaction will stop.
b) If 100 grams of calcium carbide yields 28.3 g of C2H2 g, what is the percent yield?
CaC2 (s) + 2 H2O (l) → Ca(OH)2 (aq)+ C2H2 (g)
To solve this problem, firstly we will find the GFM of CaC2 [40+(2x12) = 64 g/mole and
C2H2=[(2x12) +2x1)=26 g/mole]
According to the equation 64 grams of CaC2 produces 26 g of C2H2. The given100g of CaC2 should produce
(theoretical yield);
26 g C2H2
x g of C2H2 (theoretical yield) = 64 g CaC2 ×100 g CaC2 = 40.625 g C2H2. However, the actual yield is
only 268.3 g C2H2.
Percent error = % error =
[
[𝑀𝑒𝑎𝑠𝑢𝑟𝑒𝑑 𝑣𝑎𝑙𝑢𝑒−𝑎𝑐𝑐𝑒𝑝𝑡𝑒𝑑 𝑣𝑎𝑙𝑢𝑒]
[
𝑎𝑐𝑐𝑒𝑝𝑡𝑒𝑑 𝑣𝑎𝑙𝑢𝑒
]
[28.3 𝑔−40.625 𝑔]
40.625 𝑔
]
× 100
× 100 = 30.34%
100-36= 64. Therefore, the % yield is only 64%
To double check whether our answer is correct or not we can do the following:
28.3 𝑔
× 100 = 69.66%
40.625 𝑔
Hence, the answer 69.66% yield is correct.
22- According to the following reaction
3NO2 (g)+ H2O (l)= 2HNO3 (l)+ NO (g)
If you are given 28 g of NO2 g and 18 g of water, which reactant is acting as the limiting reactant?
To solve this problem, firstly, we will find the GFM of both reactants and then convert the given grams
of both reactants into moles.
GFM of NO2 is [ 1N +2 O= (1x14)+(2x16)= 46g/mol]
GFM of H2O is [ 2H+1 O= (2x1) + (1x16)=18 g/mol ]
mole of NO2 = 28 g of NO2 x 1mole of NO2 = 0.61 moles NO2
46 g of NO2
and moles of H2O = 18 g H2O x 1mole H2O = 1 moles H2O
18 g H2O
Secondly, we will write the calculated moles ratio right below the respective reactants in the given
equation.
3NO2 (g) + H2O (l)= 2HNO3 (l)+ NO (g)
0.61 mol
1 mol
Per the given balanced chemical equation 1 mole of H2O reacts with 3 moles of NO2 (g). However, the
calculated ratio of moles clearly shows that for 1 mole of H2O there are 0.61 moles of NO2 (g)
available, which are way less than what the reaction needs. Therefore, it is obvious that the NO2 (g) is
the limiting reactant because after its consumption the chemical reaction will stop.