# $ P{H_3} + {O_2} \to {P_2}{O_5} + {H_2}O $

The above equation is unbalanced and coefficients are the lowest whole number, find the coefficient of $ {H_2}O $

$ A. 1 \\

B. 2 \\

C. 3 \\

D. 4 \\

E. 5 \\ $

Last updated date: 17th Mar 2023

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Answer

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**Hint :**in a balanced chemical equation each element involved in the reaction has an equal number of atoms both on the reactant as well as product side. Every chemical equation needs to be balanced to be consistent with the law of conservation of matter.

**Complete Step By Step Answer:**

This particular equation given is unbalanced so first, we have to balance it. So for balancing any chemical equation we can follow the following steps:

Count the number of atoms of each element on both the product and the reactant side.

Determine which atoms are not balanced

Balance one atom at a time, using coefficients.

Start with atoms that appear only once in the reactants and only once in the products. Usually leave H atoms followed by Oxygen atoms until last.

After you are sure that you have successfully balanced the equation, repeat step 1, to be sure that mass conservation has been achieved.

Now using the following steps let us try to balance the following equation: $ P{H_3} + {O_2} \to {P_2}{O_5} + {H_2}O $

We will call the reactant side LHS and product side RHS. We will first balance the P atom. On RHS $ 2 $ P atoms are present so we will write $ 2 $ on LHS to balance it.

$ 2P{H_3} + {O_2} \to {P_2}{O_5} + {H_2}O $

Next we will balance H atom, on LHS there are $ 6 $ H atoms and on RHS only $ 2 $ so we will write $ 3 $ in front of it to balance

$ 2P{H_3} + {O_2} \to {P_2}{O_5} + 3{H_2}O $

Then, at last, we count the total number of oxygen atoms on RHS is $ 8 $ and LHS has only $ 2 $ Oxygen atoms so we will write $ 4 $ in front of it to balance.

$ 2P{H_3} + 4{O_2} \to {P_2}{O_5} + 3{H_2}O $

Hence our above equation is balance.

**Therefore the coefficient of $ {H_2}O $ is option $ C.{\text{ }}3 $ .**

**Note :**

Coefficients represent the total number of atoms present in that particular element. And they are written in front of every element. These are always in the whole number. Coefficients and subscripts are different; we never add coefficients and subscripts together.

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