Question

part 3 (1 point)
remember that chemists typically do not include a coefficient of \1\ in a balanced equation. if no coefficient is entered, it is understood that it is a \1.\
now balance the equation by placing whole - number coefficients in front of the relevant formulas.
xef6(l)+ h2o(l)→ xeo3(s)+ hf(g)

Explanation:

Step1: Balance Xe atoms

The number of Xe atoms on the left - hand side in $XeF_6$ is 1. On the right - hand side in $XeO_3$, there is 1 Xe atom. So, the coefficient of $XeF_6$ and $XeO_3$ is 1 for now.
$1XeF_6(l)+H_2O(l)\longrightarrow1XeO_3(s) + HF(g)$

Step2: Balance F atoms

In $XeF_6$, there are 6 F atoms. In HF, to balance the F atoms, the coefficient of HF should be 6.
$1XeF_6(l)+H_2O(l)\longrightarrow1XeO_3(s)+6HF(g)$

Step3: Balance H and O atoms

On the right - hand side, there are 6 H atoms in 6HF and 3 O atoms in $XeO_3$. On the left - hand side, to balance the H and O atoms, the coefficient of $H_2O$ should be 3.
$XeF_6(l)+3H_2O(l)\longrightarrow XeO_3(s)+6HF(g)$

Answer:

1; 3; 1; 6