1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Marina86 [1]
3 years ago
10

A mixture of 10cm3 of o2 and 50 cm3 of hydrogen gas sparked together to form water calculate volume of non limiting reagent left

after reacton completion
Chemistry
1 answer:
Firdavs [7]3 years ago
6 0

Answer:

30 cm³.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

O2 + 2H2 —> 2H2O

From the balanced equation above, we can say that:

1 cm³ of O2 reacted with 2 cm³ of H2 to produce 2 cm³ of H2O.

Next, we shall determine the excess reactant.

This can be obtained as follow:

From the balanced equation above,

1 cm³ of O2 reacted with 2 cm³ of H2.

Therefore, 10 cm³ of O2 will react with = (10 × 2)/1 = 10 × 2 = 20 cm³ of H2.

From the calculations made above, we can see that only 20 cm³ out of 50 cm³ of H2 given is needed to react completely with 10 cm³ of O2.

Therefore, O2 is the limiting reactant and H2 is the excess reactant (non limiting reactant).

Finally, we shall determine the volume of the non limiting reactant (excess react) that is remaining after the reaction.

This can be obtained as follow:

Volume of non limiting reactant (H2) = 50 cm³

Volume of non limiting reactant (H2) that reacted = 20 cm³

Volume of non limiting reactant (H2) remaining =.?

Volume of non limiting reactant (H2) remaining = (Volume of non limiting reactant) – (Volume of non limiting reactant that reacted)

Volume of non limiting reactant (H2) remaining = 50 – 20

= 30 cm³

Therefore, the Volume of non limiting reactant (H2) that remains after the reaction is 30 cm³

You might be interested in
An adult mosquitoes lifespan is about 16 days how many minutes is that
Delicious77 [7]
1 day = 1, 440 minutes
16 days = (1, 440 × 16) days = 23, 040 minutes
5 0
3 years ago
When a solute is added to a solution why does it remain homogeneous?
Olenka [21]
<h3>Answer:</h3>

                 When a solute is added to a solution, it remains homogeneous because the solute is soluble in given solvent.

<h3>Explanation:</h3>

                         Homogeneous mixtures, also called true solutions are those mixtures in which the components proportions are same throughout in any given sample. For example, the mixture of table salt (NaCl) and water. When the solution is unsaturated and further NaCl is added to it, it will dissolve the NaCl because the saturation point is still not reached. Remember, as "<em>Like Dissolves Like</em>" NaCl being polar in nature will interact with water molecules and will dissociate into Na⁺ and Cl⁻ ions surrounded by δ- O and δ+ H atoms of water molecules.

<h3>Conclusion:</h3>

                  In order to form a Homogeneous mixture the solution must be unsaturated, solvent must have affinity for incoming solute particles and the size of solute should be equal to 1 Â (Angstrom).

8 0
3 years ago
Read 2 more answers
What problems may global warming cause?
koban [17]

Answer:

Global warming stresses ecosystems through temperature rises, water shortages, increased fire threats, drought, weed and pest invasions, intense storm damage and salt invasion, just to name a few.

Explanation:

Global warming stresses ecosystems through temperature rises, water shortages, increased fire threats, drought, weed and pest invasions, intense storm damage and salt invasion, just to name a few.

6 0
3 years ago
H2(g) + ? O2(g) ? H2O(g) what is the correct coefficient for hydrogen gas?
Orlov [11]

Explanation:

02 is the correct coefficient for hydrogen gas .

7 0
3 years ago
4 Laconcentracion delcado clorhiclrico comercial
Darya [45]
EJERCICIOS RESUELTOS DISOLUCIONES
1.- Se disuelven 20 = g de NaOH en 560 g de agua. Calcula a) la concentración de la disolución en % en masa b) su molalidad.
Ar(Na) 23. Ar(O)=16. Ar(H)=1.
1mol NaOH = X ;X =0,5moles. m= moles(soluto) ; m= 0,5moles =0,89m; 40 g 20 g m(kg) de disolvente 0,56 kg
2.- ¿Qué cantidad de glucosa, C6H12O6 (Mm = 180 g/mol), se necesita para preparar 100 cm3 de disolución 0,2 molar?
M = moles(soluto) ; moles C6 H12O6 = M.V = 0,2M.0,1l; moles C6 H12O6 = 0,02. V (l) de disolución
1 mol glu cos a = 0,02 moles ; X = 36 g. 180 g X
3.- Se dispone de un ácido nítrico comercial concentrado al 96,73 % en masa y densidad 1,5 g/mL. ¿Cuántos mL del ácido concentrado serán necesarios para preparar 0,2 L de disolución 1,5 M de dicho ácido? Mm (HNO3) = 63g/mol.
Primeramente calcularemos los moles de ácido puro que necesitamos:
M= moles(soluto) ; moles(HNO3)=M.V=1,5M.0,2l=0,3. V (l) de disolución
Ahora calculamos la masa en g correspondiente:
0,3moles x 63g = 18,9 g de HNO3 . 1mol
Como el ácido comercial del que disponemos no es puro, sino del 96,73 % necesitaremos pesar:
100g del ácido comercial = X ; X =19,54g ácido comercial. contienen 96,73g ácido puro 18,9g ácido puro
Como necesitamos averiguar el volumen en mL que hemos de coger, utilizamos la densidad del ácido comercial:
d(g/ml)= m(g) ; V(ml)= 19,54g =13ml. V ( m l ) 1, 5 g / m l
a)
%NaOH = m(g)NaOH .100; m( g )disolución
%NaOH = 20 .100; 580
%NaOH = 3,45. b) Primeramente calculamos los moles que son los 20 g de soluto:

4.- Calcula la masa de nitrato de hierro (II), Fe(NO3)2, que hay en 100 mL de disolución acuosa al 6 %. Densidad de la disolución 1,16 g/mL
De la densidad sabemos que los 100 ml de disolución tienen de masa 116 g. Como es al 6 %, la masa de soluto existente será:
En 100g disolución = En 116g disolución ; X = 6,96g Fe(NO3 )2 . hay 6g Fe(NO3 )2 X
5.- Indica de qué modo prepararías 1⁄2 L de disolución 0,1 M de HCl si disponemos de un HCl concentrado del 36 % y densidad 1,19 g/mL
Calculamos la masa de HCl que necesitamos. Para ello, utilizando el concepto de molaridad, averiguamos primeramente los moles de HCl que va a tener la disolución que queremos preparar:
n(HCl) = M.V = 0,1M.0,5l = 0,05moles.
C o m o M m ( H C l ) = 3 6 , 5 g / m o l . L o s 0 , 0 5 m o l e s s e r á n : 0 , 0 5 m o l e s . 3 6 , 5 g = 1, 8 3 g H C l .
1mol
Esa masa de HCl la tenemos que coger del HCl concentrado del que se dispone (36 % y densidad 1,19
g/ml.). Al no ser puro, sino del 36 % tendremos que coger más cantidad de gramos: 100g del HCl concentrado = X ; X = 5,08g HCl puro.
contienen 36g HCl puro 1,83g HCl puro
Como se trata de un líquido del que conocemos su densidad, determinamos el volumen de esos 5,08 g:
V=m; V= 5,08g =4,27mlHCldel36%. ρ 1,19g / ml
Preparación: En un matraz aforado de 1⁄2 l que contenga algo de agua destilada, se introducen 4,27 ml del HCl concentrado del 36 %, utilizando una pipeta. No absorber el ácido con la boca porque es tóxico.
Se agita con cuidado el matraz hasta que se disuelva el soluto.
Se añade agua destilada al matraz hasta alcanzar exactamente la señal de 500 ml.
6.- Se disuelven en agua 30,5 g de cloruro amónico (NH4Cl) hasta obtener 0,5 l de disolución. Sabiendo que la densidad de la misma es 1027 kg/m3, calcula:
a) La concentración de la misma en porcentaje en masa. b) La molaridad.
c) La molalidad.
d) Las fracciones molares del soluto y del disolvente.
Mm(NH4Cl)=53,5g/mol.
Primeramente 1027kg/m3 = 1,027 g/cm3. Luego la masa de 1 l de disolución será de 1027 g y la de medio litro 513,8 g. De ellos 30,5 g son de soluto (cloruro amónico) y el resto 483,3 g son de agua.

a) %masa NH4Cl = masa(g)soluto x100 = 30,5g x100 = 5,94%. masa(g)disolución 513,8g
b) M = moles soluto =
Planteamos la ecuación con los moles de manera que la suma de los que tomamos de la disolución A más los que tomamos de la disolución B sea igual a 0,5
1,5.V +0,1(1−V )=0,5; V =0,286l=286cm3. V =0,714l=714cm3. AAA B
8 0
3 years ago
Other questions:
  • Professional scientific "magazines" in which scientists write
    13·1 answer
  • K2so4 is a strong electrolyte. determine the concentration of each of the individual ions in a 0.450 m k2so4 solution.
    11·1 answer
  • A 1.00-degree increase on the Celsius scale is equivalent to a 1.80-degree increase on the Fahrenheit scale. The temperature of
    12·2 answers
  • The compound known as butylated hydroxytoluene, abbreviated as BHT, contains carbon, hydrogen, and oxygen. A 2.001 g sample of B
    6·1 answer
  • A reaction has ∆H = −356 kJ and ∆S = −36 J/K. Calculate ∆G (kJ) at 25°C.
    5·1 answer
  • I’m doing a project in school on liquids and gasses and I wanted to know if air conditioning is a gas?
    9·1 answer
  • Write the word and balanced chemical equations for the reaction between:
    13·1 answer
  • When barium becomes positively charged, it forms a(n): (2 points)
    12·1 answer
  • Calculate the mass percentage of CaCO3 in the 103g mixture of CaCO3 and Na2CO3, which produces 24 litres of CO2 at one atmospher
    14·1 answer
  • you wish to prepare an hc2h3o2 buffer with a ph of 5.44. if the pka of the acid is 4.74, what ratio of c2h3o2-/hc2h3o2 must you
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!