Chemistry, 04.05.2021 03:10 cathydaves

Li3N +
NH4NO3 →
LINO3 +
(NH4)3N Reaction Type:

Answers: 1

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Answer from: Emilyvite6251

Name: Class:

Five Types of Chemical Reaction Worksheet Answers

Balance the following reactions and indicate which of the six types of chemical reaction are being

represented:

a) 2 NaBr + 1 Ca(OH)2  1 CaBr2 + 2 NaOH Type of reaction: double displacement

b) 2 NH3+ 1 H2SO4  1 (NH4)2SO4 Type of reaction: synthesis

c) 4 C5H9O + 27 O2  20 CO2 + 18 H2O Type of reaction: combustion

d) 3 Pb + 2 H3PO4  3 H2 + 1 Pb3(PO4)2 Type of reaction: single displacement

e) 1 Li3N + 3 NH4NO3  3 LiNO3 + 1 (NH4)3N Type of reaction: double displacement

Section 3: Predicting the products of chemical reactions

1) 2 Ag + 1 CuSO4  1 Ag2SO4 + 1 Cu Type: Single Displacement

2) 2 NaI + 1 CaCl2  2 NaCl + 1 CaI2 Type: Double Displacement

3) 1 O2 + 1 H2  2 H2O Type: Synthesis

4) 2 AgNO2 + 1 BaSO4  1 Ag2SO4 + 1 Ba(NO2)2 Type: Double Displacement

5) 2 HCN + 1 CuSO4  1 H2SO4 + 1 Cu(CN)2 Type: Double Displacement

6) 1 H2O + 1 AgI  1 HI + 1 AgOH Type: Double Displacement

7) 4 LiBr + 1 Co(SO3)2  2 Li2SO3 + 1 CoBr4 Type: Double Displacement

8) 1 LiNO3 + 1 Ag  1 AgNO3 + 1 Li Type: Single Displacement

9) 1 N2 + 2 O2  2 NO2 Type: Synthesis

10) 1 H2CO3  1 CO2 + 1 H2O Type: Decomposition

11) 1 AlCl3 + 3 Cs  3 CsCl + 1 Al Type: Single Displacement

12) 1 Al(NO3)3 + 1 Ga  1 Ga(NO3)3 + 1 Al Type: Single Displacement

13) 1 CH3COOH + 2 O2  2 CO2 + 2 H2O Type: Combustion

14) 1 C4H8 + 6 O2  4 CO2 + 4 H2O Type: Combustion

15) 2 KCl + 1 Mg(OH)2  2 KOH + 1 MgCl2 Type: Double Displacement

16) 1 Zn + 1 Au(NO2)2  1 Zn(NO2)2 + 1 Au Type: Single Displacement

17) 1 BaS + 1 PtCl2  1 BaCl2 + 1 PtS Type: Double Displacement

18) 2 Na2O  4 Na + 1 O2 Type: Decomposition

Name: Class:

Five Types of Chemi

Explanation:

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

thats a little if u want the answer just look it up on google thats what i do haha : )

Draw the orbitals for the elements and give their number of electrons

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Li3N +
NH4NO3 →
LINO3 +
(NH4)3N Reaction Type:...

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