Question

balance the following chemical equation​

Answer 1

AnsWer :

• Balance Chemical equation define as When Reactants and Product, ( LHS and RHS ) should be equal, it said Equation be Balance.
• According to Law of Conservation of Mass Reactants always equal to Product.

$\rule{90}1$

# How to Balance Chemical Equation ?

Some Step follow to balance Chemical Equation,

Steps 1.

• Draw a table, and Count the number of atom of all the elements of Reactants and product both sides.
• Example - H + O2 ---- H2O.

$\begin{tabular}{| c | c | c |} \cline{1-3} Elements & No of atoms of Reactants & No of atoms of Product \\ \cline{1-3} H & 1 & 2 \\ O & 2 & 1 \\ \cline{1-3} \end{tabular}$

Steps 2.

• In Order to equate number of atom LHS and RHS should balance or not, If not then try to balance it.
• Multiply or divided by number of atom on both sides.
• Balance Table provide below.

$\begin{tabular}{| c | c | c |} \cline{1-3} Elements & No of atoms of Reactants & No of atoms of Product \\ \cline{1-3} H & 2 & 2 \\ O & 2 & 2 \\ \cline{1-3} \end{tabular}$

Steps 3.

• Our Equation should be balance with follow Law of Conservation of Mass.
• 2H + O2 ------ 2H2O.

$\rule{200}3$

Let Balance given chemical Equations,

23)

$\tt {3} Li + AlCl_3 \longrightarrow {3}LiCl + Al.$

$\rule{90}1$

24)

$\tt2C_4H_6 + 11 O_2 \longrightarrow 8CO_2 + 6H_2O.$

$\rule{90}1$

25)

$\tt3NH_4OH + H_3 PO_4 \longrightarrow (NO_4)3PO_4 + 3H_2O.$

$\rule{90}1$

26)

$\tt3Rb + P \longrightarrow Rb_3 P.$

$\rule{90}1$

27)

$\tt \: CH_4 + 2O_2 \longrightarrow 2 H_2O + CO_2.$

$\rule{90}1$

28)

$\tt2Al( OH )_3 + 3H_2 SO_4 \longrightarrow Al_2 (SO_4)_3 + 6H_2O.$

$\rule{90}1$

29)

$\tt2Na + Cl_2 \longrightarrow 2NaCl.$

$\rule{90}1$

30)

$\tt16 Rb + S_8 \longrightarrow 8Rb_2 S.$

$\rule{90}1$

31)

$\tt3Ca( OH )_2 + 2H_3 PO_4 \longrightarrow Ca_3( Po_4)_2 + 6H_2O.$

$\rule{90}1$

32)

$\tt NH_3 + HCl \longrightarrow NH_4Cl.$

$\rule{90}1$

33)

$\tt 2Li + 2H_2O \longrightarrow 2Li OH + H_2.$

$\rule{90}1$

34)

$\tt Ca_3 (PO_4)_2 +3SiO_2 +5C \longrightarrow 3Ca SiO_3 + 5CO + 2P.$

$\rule{90}1$

35)

$\tt4NH_3 + 3O_2 \longrightarrow 2N_2 + 6H_2O.$

$\rule{90}1$

36)

$\tt4FeS_2 + 11 O_2 \longrightarrow 2Fe_2 O_3 + 8 SO_2.$

$\rule{90}1$

37)

$\tt5C + 2SO_2 \longrightarrow CS_2 + 4CO.$

Answer 2

Balancing the Equations -

$(23.)\: \sf{\bf \underline{3}Li + AlCl_{3} \rightarrow Al}$

$(24.)\: \sf{\bf 2C_{2}H_{6}+7O_{2} \rightarrow 4CO_{2}+6H_{2}O }$

$(25.)\: \sf{\bf 3NH_{4}OH+H_{3}PO_{4} \rightarrow (NH_{4} )_{3}PO_{4}+3H_{2} O}$

$(26).\: \sf{\bf 3Rb + P \rightarrow Rb_{3} P}$

$(27).\: \sf{\bf CH_{4} + 2O_{2} \rightarrow CO_{2}+2 H_{2}O}$

$(28).\: \sf{\bf 2Al( OH )_{3} + 3H_{2} SO_{4} \rightarrow Al_{2} (SO_{4})_{3} + 6H_{2} O}$

$(29.)\: \sf{\bf 2Na + Cl_{2} \rightarrow 2NaCl}$

$(30).\: \sf{\bf 16 Rb + S_{8} \rightarrow 8Rb_{2} S}$

$(31). \: \sf{\bf 2H_{3} PO_{4} + 3Ca( OH )_{2} \rightarrow Ca_{3} ( PO_{4})_{2} + 6H_{2} O}$

$(32). \: \sf\bf {NH_{3} + HCl \rightarrow NH_{4} Cl}\:\: \underline{Balanced \:\:already}$

$(33).\: \sf{\bf 2Li + 2H_{2} O \rightarrow 2Li OH + H_{2}}$

$(34).\: \sf{ \bf Ca_{3} (PO_{4})_{2} +3SiO_{2} +5C \rightarrow 3Ca SiO_{3} + 5CO + 2P}$

$(35).\: \sf{\bf 4NH_{3} + 3O_{2} \rightarrow 2N_{2} + 6H_{2} O}$

$(36). \: \sf{\bf 4FeS_{2} + 11 O_{2} \rightarrow 2Fe_{2} O_{3} + 8 SO_{2}}$

$(37).\: \sf{\bf 5C + 2SO_{2} \rightarrow CS_{2} + 4CO}$