All Solutions
Section 6-4: Information in Chemical Equations
$$
mathrm{C_3H_{8(g)} + O_{2(g)} longrightarrow CO_{2(g)} + H_2O_{(g)}}
$$
But its balanced chemical equation is:
$$
mathrm{C_3H_{8(g)} + 5O_{2(g)} longrightarrow 3CO_{2(g)} + 4H_2O_{(g)}}
$$
$$
mathrm{NaOH_{(aq)} + HCl_{(aq)} longrightarrow NaCl_{(aq)} + H_2O_{(l)}}
$$
The correct balanced equation is :
$$
mathrm{2HI_{(g)} longrightarrow H_{2(g)} + I_{2(g)}}
$$
This is because for balancing equation, the rule is to only add whole number constants in on the left hand side of the element or molecule. Adding constants in subscripts is not allowed as the subscripts represent the combining ratio of the elements in a compound which is fixed and not variable.
Consider the equation:
$$
mathrm{2HI_{(g)} longrightarrow H_{2(g)} + I_{2(g)}}
$$
Here the 2 on the left hand side of HI is the coefficient of HI in this equation, while subscripts 2 on the lower right hand side of both hydrogen gas H and Iodine I are subscripts. The subscripts represent the combining ratio of elements in a compound. For example H$_2$ means that 2 atoms of hydrogen form one molecule of hydrogen gas.
Balancing of chemical equations can only be done by changing the coefficients of reactant(s) ad product(s) and not the subscripts. Adding constants in subscripts is not allowed as the subscripts represent the combining ratio of the elements in a compound which is fixed and not variable.
$$
mathrm{2KI_{(aq)} longrightarrow 2K_{(s)} + I_{2(l)}}
$$
$$
mathrm{Mg_{(s)}+2AgNO_{3(aq)} longrightarrow Mg(NO_{3})_{2(aq)} + 2Ag_{(s)}}
$$
$$
mathrm{2Na_{(s)}+2H_2O{(aq)} longrightarrow 2NaOH_{(aq)} + H_{2(g)}}
$$
$$
mathrm{Pb(NO_{3})_{2(aq)}+2NaCl_{(aq)} longrightarrow PbCl_{2(aq)} + 2NaNO_{3(aq)}}
$$
$$
mathrm{2C_8H_{18(l)} + 25O_{2(g)} longrightarrow 16CO_{2(g)} + 18H_2O_{(g)}}
$$
2 molecules of octane produce 16 molecules of carbon dioxide on complete combustion. This implies that 1 molecule of octane will produce $dfrac{16}{2}=8$ molecules of carbon dioxide.
$$
mathrm{Ca+Cl_2 longrightarrow CaCl_{2}}
$$
$$
mathrm{2K+Br_2 longrightarrow 2KBr}
$$
$$
mathrm{2H_2O_2 longrightarrow 2H_{2}O+ O_2}
$$
$$
mathrm{4Na+O_2 longrightarrow 2Na_{2}O}
$$
$$
mathrm{N_2+3H_2 longrightarrow 2NH_{3}}
$$
$$
mathrm{NH_4OH+HBr longrightarrow H_{2}O+NH_4Br}
$$
$$
mathrm{CaSO_4+2KOH longrightarrow Ca(OH)_{2}+K_2SO_4}
$$
$$
mathrm{Ba+2HNO_3 longrightarrow Ba(NO_3)_{2}+H_2}
$$
$$
mathrm{H_3PO_4+3NaOH longrightarrow Na_3PO_4+3H_{2}O}
$$
$$
mathrm{C_3H_8 + 5O_{2} longrightarrow 3CO_{2} + 4H_2O}
$$
$$
mathrm{Al_4C_3+12H_2O longrightarrow 3CH_4+4Al(OH)_{3}}
$$
$$
mathrm{FeBr_3+3Na longrightarrow Fe+3NaBr}
$$
$$
mathrm{2Fe+3H_2SO_4longrightarrow Fe_2(SO_4)_3+3H_{2}}
$$
$$
mathrm{2C_2H_6 + 7O_{2} longrightarrow 4CO_{2} + 6H_2O}
$$
Word equation:
$$
text{ammonium dichromate} longrightarrowtext{nitrogen gas + water vapour}
$$
According to the law of conservation of masses, the total mass of the products of a reaction is equal and total mass of its reactants. This implies that if 2.5 grams of ammonium dichromate decomposed to release 1 gram of nitrogen gas and water vapour, then $2.5-1=1.5$ grams of the solid product was formed.
b. 1.5 grams of solid product was formed.
