The percentage of \[{P_2}{O_3}\] in \[{H_3}P{O_3}\] is:
A. 44.69
B. 33.52
C. 67.07
D. 50.04
Answer
583.8k+ views
Hint:To calculate the percentage of phosphorus trioxide in phosphorus acid find out the mass of each compound by the formula of moles where number of moles is equal to mass divided by volume. The moles can be determined by the balanced chemical equation.
Complete step by step answer:
The phosphorous acid is formed by reacting phosphorus trioxide with water.
\[{P_2}{O_3} + 3{H_2}O \to 2{H_3}P{O_3}\]
In this reaction, one mole of phosphorus trioxide reacts with three mole of water to form two mole of phosphorus acid.
The formula to calculate the number of moles is given below.
\[n = \dfrac{m}{M}\]
n is number of moles.
m is given mass.
M is molecular weight
The molecular weight of phosphorous acid is 81.996 g/mol.
To calculate the mass of phosphorous acid, substitute the values in the above equation.
\[ \Rightarrow 2 = \dfrac{m}{{81.996g/mol}}\]
\[ \Rightarrow m = 2 \times 81.996\]
\[ \Rightarrow m = 163.992g\]
The molecular weight of phosphorus trioxide is 109.946 g/mol.
To calculate the mass of phosphorus trioxide, substitute the value in the above equation.
\[ \Rightarrow 1 = \dfrac{m}{{109.946g/mol}}\]
\[ \Rightarrow m = 109.946g\]
The mass % of \[{P_2}{O_3}\] is calculated by the formula as shown below.
\[\% A = \dfrac{{{m_A}}}{{{m_B}}} \times 100\]
Where,
\[{m_A}\] is the mass of compound A.
\[{m_B}\] is the mass of compound B.
To find the percentage of \[{P_2}{O_3}\] in \[{H_3}P{O_3}\], substitute the values in the above equation.
\[ \Rightarrow \% {P_2}{O_3} = \dfrac{{110}}{{164}} \times 100\]
\[ \Rightarrow \% {P_2}{O_3} = 67.07\% \]
Thus, the percentage of \[{P_2}{O_3}\] in \[{H_3}P{O_3}\ ] is 67.07 %.
Therefore, the correct option is C.
Note:
Make sure to round off the values of mass obtained to calculate the percentage of phosphorus trioxide. For calculating the mass of solute present in solution, mass of solute is divided by the mass of solution multiplied by 100.
Complete step by step answer:
The phosphorous acid is formed by reacting phosphorus trioxide with water.
\[{P_2}{O_3} + 3{H_2}O \to 2{H_3}P{O_3}\]
In this reaction, one mole of phosphorus trioxide reacts with three mole of water to form two mole of phosphorus acid.
The formula to calculate the number of moles is given below.
\[n = \dfrac{m}{M}\]
n is number of moles.
m is given mass.
M is molecular weight
The molecular weight of phosphorous acid is 81.996 g/mol.
To calculate the mass of phosphorous acid, substitute the values in the above equation.
\[ \Rightarrow 2 = \dfrac{m}{{81.996g/mol}}\]
\[ \Rightarrow m = 2 \times 81.996\]
\[ \Rightarrow m = 163.992g\]
The molecular weight of phosphorus trioxide is 109.946 g/mol.
To calculate the mass of phosphorus trioxide, substitute the value in the above equation.
\[ \Rightarrow 1 = \dfrac{m}{{109.946g/mol}}\]
\[ \Rightarrow m = 109.946g\]
The mass % of \[{P_2}{O_3}\] is calculated by the formula as shown below.
\[\% A = \dfrac{{{m_A}}}{{{m_B}}} \times 100\]
Where,
\[{m_A}\] is the mass of compound A.
\[{m_B}\] is the mass of compound B.
To find the percentage of \[{P_2}{O_3}\] in \[{H_3}P{O_3}\], substitute the values in the above equation.
\[ \Rightarrow \% {P_2}{O_3} = \dfrac{{110}}{{164}} \times 100\]
\[ \Rightarrow \% {P_2}{O_3} = 67.07\% \]
Thus, the percentage of \[{P_2}{O_3}\] in \[{H_3}P{O_3}\ ] is 67.07 %.
Therefore, the correct option is C.
Note:
Make sure to round off the values of mass obtained to calculate the percentage of phosphorus trioxide. For calculating the mass of solute present in solution, mass of solute is divided by the mass of solution multiplied by 100.
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