Solution:- (16) [tex]P_4(s)+F_2(g)\rightarrow PF_3(s)[/tex]
For this equation, we multiply the product side by 4 to balance phosphorous. On doing this, there becomes 12F on product side so to balance F we need to multiply the fluorine on reactant side by 6. Since both sides have four P, we multiply the reactant side phosphorous by just 1 to fill in the blank space.
[tex]P_4(s)+6F_2(g)\rightarrow 4PF_3(s)[/tex]
(17) The given equation is:
[tex]C(s)+H_2O(g)\rightarrow CH_4(g)+CO_2(g)[/tex]
Product side we have total 2 carbons, so to make 2 carbons on reactant side we multiply the carbon by 2.
Product side has four hydrogen and two oxygen and the reactant side has two hydrogen and one oxygen. So, to balance hydrogen and oxygen we multiply reactant side water by 2. The balanced equation looks as:
[tex]2C(s)+2H_2O(g)\rightarrow CH_4(g)+CO_2(g)[/tex]
(18) The given equation is:
[tex]HgO(s)\rightarrow O_2(g)+Hg(l)[/tex]
In this equation, both sides have equal Hg but the oxygen is not balanced as the reactant side has one O and the product side has two O. To balance oxygen we multiply the reactant side by 2. On doing this, there becomes two Hg on reactant side. So, to makes two Hg on product side also we multiply Hg by 2. The balanced equation looks as:
[tex]2HgO(s)\rightarrow O_2(g)+2Hg(l)[/tex]
(19) The given equation is:
[tex]CaCO_3(s)\rightarrow CO_2(g)+CaO(s)[/tex]
This equation has one Ca, one C and three O on both sides so it is already balanced. so, just put 1 to each blank box.
[tex]CaCO_3(s)\rightarrow CO_2(g)+CaO(s)[/tex]
The below image shows the filled boxes:
