All categories

2 years ago

Use the unbalanced equation NH3 + O2 = NO + H2O, what is the mole ratio between NH3 and NO

Chemistry

2 answers:

vovangra [49]2 years ago

5 0

Answer: The mole ratio between $NH_3$ and $NO$ is, 1 : 1

Explanation:

The given unbalanced chemical reaction is,

$NH_3+O_2\rightarrow NO+H_2O$

This is an unbalanced chemical reaction because in this reaction the number oxygen and hydrogen atoms are not balanced on both side.

In order to balance the chemical reaction, the coefficient 4 is put before $NH_3$ , coefficient 3 is put before $O_2$ , coefficient 4 is put before $NO$ and coefficient 2 is put before $H_2O$ .

The balanced chemical reaction will be,

$4NH_3+3O_2\rightarrow 4NO+2H_2O$

From the balanced chemical reaction we conclude that,

The mole ratio of $NH_3:O_2:NO:H_2O$ are 4 : 3 : 4 : 2

The mole ratio of $NH_3:NO$ are 4 : 4 or 1 : 1

Therefore, the mole ratio between $NH_3$ and $NO$ is, 1 : 1

kirill115 [55]2 years ago

4 0

The first thing that needs to be done in order to answer the question above is to balance first the chemical equation by seeing to that the number of moles of a certain element on the reactant side is equal to the number of moles in the product side. 4NH3 + 5O2  4NO + 6H2O The mole fraction between the NH3 and NO is therefore 4:4 or 1:1.

You might be interested in

1.2 moles of (NH4)3PO3

Aleks04 [339]

1.2 moles of (nph4)3po3 is.......159.6 grams

3 0

2 years ago

An unknown metal absorbs 61 J of heat, and its temperature increases by 29∘C. What is the heat capacity of the metal?

aniked [119]

Answer:

2.103 J/C

Explanation:

Quantity of heat = Heat Capacity * Temperature change

Heat Capacity = Quantity of heat / Temperature Change

Heat Capacity = 61/29

Heat Capacity = 2.103 J/C

3 0

3 years ago

Identify reagents necessary to convert cyclohexane into 1,3-cylohexadiene: notice that the starting material has no leaving grou

Vinvika [58]

1,3-cylohexadiene i synthesized starting from cyclohexane in following 4 steps.

1) Free Radical Substitution Rxn: Halogenation of cyclohexane in the presence of UV yield chlorocyclohexane.

2) Elimination Rxn: Dehydrohalogenation of chlorocyclohexane yields cyclohexene.

3) Halogenation of Cyclohexene (Electrophillic Addition Rxn) gives 1,2-dihalocyclohexane.

4) Elemination Rxn: When dibromocyclohexane is treated with KOH and heated it gives 1,3-cyclohexadiene as shown below,

4 0

2 years ago

I NEED HELP PLEASE!!!! CHEMISTRY QUESTION: If 38 g of Li3P and 15 grams of Al2O3 are reacted, what total mass of products will r

maksim [4K]

Answer:

21.5 g.

Explanation:

Hello!

In this case, since the reaction between the given compounds is:

$2Li_3P+Al_2O_3\rightarrow 3Li_2O+2AlP$

We can see that according to the law of conservation of mass, which states that matter is neither created nor destroyed during a chemical reaction, the total mass of products equals the total mass of reactants based on the stoichiometric proportions; in such a way, we first need to compute the reacted moles of Li3P as shown below:

$n_{Li_3P}^{reacted}=38gLi_3P*\frac{1molLi_3P}{51.8gLi_3P}=0.73molLi_3P$

Now, the moles of Li3P consumed by 15 g of Al2O3:

$n_{Li_3P}^{consumed \ by \ Al_2O_3}=15gAl_2O_3*\frac{1molAl_2O_3}{101.96gAl_2O_3} *\frac{2molLi_3P}{1molAl_2O_3} =0.29molLi_3P$

Thus, we infer that just 0.29 moles of 0.73 react to form products; which means that the mass of formed products is:

$m_{Li_2O}=0.29molLi_3P*\frac{3molLi_2O}{2molLi_3P} *\frac{29.88gLi_2O}{1molLi_2O} =13gLi_2O\\\\m_{AlP}=0.29molLi_3P*\frac{2molAlP}{2molLi_3P} *\frac{57.95gAlP}{1molAlP} =8.5gAlP$

Therefore, the total mass of products is:

$m_{products}=13g+8.5g\\\\m_{products}=21.5g$

Which is not the same to the reactants (53 g) because there is an excess of Li₃P.

Best Regards!

7 0

2 years ago

Calculate the empirical formula for each stimulant based on its elemental mass percent composition. a. nicotine (found in tobacc

Ira Lisetskai [31]

This an incomplete question, here is a complete question.

Calculate the empirical formula for each stimulant based on its elemental mass percent composition.

a. nicotine (found in tobacco leaves): C 74.03%, H 8.70%, N 17.27%

b. caffeine (found in coffee beans): C 49.48%, H 5.19 %, N 28.85% and O 16.48%

Answer:

(a) The empirical formula for the given compound is $C_5H_7N$

(b) The empirical formula for the given compound is $C_4H_5N_2O$

Explanation:

Part A: nicotine

We are given:

Percentage of C = 74.03 %

Percentage of H = 8.70 %

Percentage of N = 17.27 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 74.03 g

Mass of H = 8.70 g

Mass of N = 17.27 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{74.03g}{12g/mole}=6.17moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{8.70g}{1g/mole}=8.70moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{17.27g}{14g/mole}=1.23moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.23 moles.

For Carbon = $\frac{6.17}{1.23}=5.01\approx 5$

For Hydrogen = $\frac{8.70}{1.23}=7.07\approx 7$

For Nitrogen = $\frac{1.23}{1.23}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N = 5 : 7 : 1

The empirical formula for the given compound is $C_5H_7N_1=C_5H_7N$

Part B: caffeine

We are given:

Percentage of C = 49.48 %

Percentage of H = 5.19 %

Percentage of N = 28.85 %

Percentage of O = 16.48 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 49.48 g

Mass of H = 5.19 g

Mass of N = 28.85 g

Mass of O = 16.48 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{49.48g}{12g/mole}=4.12moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{5.19g}{1g/mole}=5.19moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{28.85g}{14g/mole}=2.06moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{16.48g}{16g/mole}=1.03moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.03 moles.

For Carbon = $\frac{4.12}{1.03}=4$

For Hydrogen = $\frac{5.19}{1.03}=5.03\approx 5$

For Nitrogen = $\frac{2.06}{1.03}=2$

For Nitrogen = $\frac{1.03}{1.03}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N : O = 4 : 5 : 2 : 1

The empirical formula for the given compound is $C_4H_5N_2O_1=C_4H_5N_2O$

6 0

3 years ago