Chemistry, 13.06.2020 22:57 giavanleer14

balance the following equations show the number of atoms for each element in each side of the equation

balance the following equations show the number of atoms for each element in each side of the equat

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Answer from: alysonmariefont

Answer :

(1) $Cl_2+2LiI\rightarrow 2LiCl+I_2$

(2) $N_2+3H_2\rightarrow 2NH_3$

(3) $Ca(OH)_2+2HCl\rightarrow CaCl_2+H_2O$

(4) $2Al(NO_3)_3+3H_2SO_4\rightarrow Al_2(SO_4)_3+6HNO_3$

(5) $C_4H_8+6O_2\rightarrow 4CO_2+4H_2O$

(6) $(NH_4)_2SO_4+2KOH\rightarrow K_2SO_4+2NH_3+2H_2O$

(7) $2C_2H_2+5O_2\rightarrow 4CO_2+2H_2O$

(8) $2H_3PO_4+3Mg(OH)_2\rightarrow Mg_3(PO_4)_2+6H_2O$

(9) $Al(NO_3)_3+3NaOH\rightarrow 3NaNO_3+Al(OH)_3$

(10) $BaCl_2+Na_2SO_4\rightarrow BaSO_4+2NaCl$

Explanation :

Balanced chemical reaction : It is defined as the reaction in which the number of atoms of individual elements present on reactant side must be equal to the product side.

Part 1 :

The unbalanced chemical reaction is,

$Cl_2+LiI\rightarrow LiCl+I_2$

In order to balance the chemical equation, the coefficient '2' put before the LiI and LiCl then we get the balanced chemical equation.

The balanced chemical reaction will be,

$Cl_2+2LiI\rightarrow 2LiCl+I_2$

Part 2 :

The unbalanced chemical reaction is,

$N_2+H_2\rightarrow NH_3$

In order to balance the chemical equation, the coefficient '3' put before the H_2 and the coefficient '2' put before the NH_3 then we get the balanced chemical equation.

The balanced chemical reaction will be,

$N_2+3H_2\rightarrow 2NH_3$

Part 3 :

The unbalanced chemical reaction is,

$Ca(OH)_2+HCl\rightarrow CaCl_2+H_2O$

In order to balance the chemical equation, the coefficient '2' put before the HCl then we get the balanced chemical equation.

The balanced chemical reaction will be,

$Ca(OH)_2+2HCl\rightarrow CaCl_2+H_2O$

Part 4 :

The unbalanced chemical reaction is,

$Al(NO_3)_3+H_2SO_4\rightarrow Al_2(SO_4)_3+HNO_3$

In order to balance the chemical equation, the coefficient '2' put before the Al(NO_3)_3 , the coefficient '3' put before the H_2SO_4 and the coefficient '6' put before the HNO_3 then we get the balanced chemical equation.

The balanced chemical reaction will be,

$2Al(NO_3)_3+3H_2SO_4\rightarrow Al_2(SO_4)_3+6HNO_3$

Part 5 :

The unbalanced chemical reaction is,

$C_4H_8+O_2\rightarrow CO_2+H_2O$

In order to balance the chemical equation, the coefficient '6' put before the O_2 , the coefficient '4' put before the CO_2 and the coefficient '4' put before the H_2O then we get the balanced chemical equation.

The balanced chemical reaction will be,

$C_4H_8+6O_2\rightarrow 4CO_2+4H_2O$

Part 6 :

The unbalanced chemical reaction is,

$(NH_4)_2SO_4+KOH\rightarrow K_2SO_4+NH_3+H_2O$

In order to balance the chemical equation, the coefficient '2' put before the KOH,NH_3 and H_2O then we get the balanced chemical equation.

The balanced chemical reaction will be,

$(NH_4)_2SO_4+2KOH\rightarrow K_2SO_4+2NH_3+2H_2O$

Part 7 :

The unbalanced chemical reaction is,

$C_2H_2+O_2\rightarrow CO_2+H_2O$

In order to balance the chemical equation,the coefficient '2' put before the C_2H_2 , the coefficient '5' put before the O_2 , the coefficient '4' put before the CO_2 and the coefficient '2' put before the H_2O then we get the balanced chemical equation.

The balanced chemical reaction will be,

$2C_2H_2+5O_2\rightarrow 4CO_2+2H_2O$

Part 8 :

The unbalanced chemical reaction is,

$H_3PO_4+Mg(OH)_2\rightarrow Mg_3(PO_4)_2+H_2O$

In order to balance the chemical equation,the coefficient '2' put before the H_3PO_4 , the coefficient '3' put before the Mg(OH)_2 and the coefficient '6' put before the H_2O then we get the balanced chemical equation.

The balanced chemical reaction will be,

$2H_3PO_4+3Mg(OH)_2\rightarrow Mg_3(PO_4)_2+6H_2O$

Part 9 :

The unbalanced chemical reaction is,

$Al(NO_3)_3+NaOH\rightarrow NaNO_3+Al(OH)_3$

In order to balance the chemical equation,the coefficient '3' put before the NaOH and the coefficient '3' put before the NaNO_3 then we get the balanced chemical equation.

The balanced chemical reaction will be,

$Al(NO_3)_3+3NaOH\rightarrow 3NaNO_3+Al(OH)_3$

Part 10 :

The unbalanced chemical reaction is,

$BaCl_2+Na_2SO_4\rightarrow BaSO_4+NaCl$

In order to balance the chemical equation,the coefficient '2' put before the NaCl then we get the balanced chemical equation.

The balanced chemical reaction will be,

$BaCl_2+Na_2SO_4\rightarrow BaSO_4+2NaCl$

Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

Answer from: Quest

C: plants and animals

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