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jeyben [28]
3 years ago
12

Zinc metal reacts with hydrochloric acid to produce zinc(II) chloride and hydrogen gas. How many liters of hydrogen gas will be

produced at STP when 10.00 g of zinc metal is added to 23.8 mL of a 0.45 M hydrochloric acid solution?
Chemistry
1 answer:
AleksAgata [21]3 years ago
4 0

Answer:

0.120 L of hydrogen gas will be produced

Explanation:

Step 1: Data given

Mass of zinc = 10.0 grams

Volume of hydrochloric acid = 23.8 mL

Molarity of hydrochloric acid = 0.45 M

Molar mass of zinc =65.38 g/mol

Step 2: The balanced equation

Zn + 2HCl → ZnCl2 + H2

Step 3: Calculate moles Zinc

Moles Zn = mass Zn / molar mass Zn

Moles Zn = 10.0 grams / 65.38 g/mol

Moles Zn  =  0.153 moles

Step 4: Calculate moles HCl

Moles HCl = molarity * volume

Moles HCl = 0.45 M * 0.0238 L

Moles HCl = 0.01071 moles

Step 5: Calculate limiting reactant

For 1 mol Zn, we need 2 moles HCl to produce 1 mol ZnCl2 and 1 mol H2

HCl is the limiting reactant. It will completely be consumed (0.01071 moles)

Zn is in excess. There will react 0.01071/2 = 0.005355 moles

There will remain 0.153 - 0.005355 = 0.147645 moles

Step 6: Calculate moles H2

For 1 mol Zn, we need 2 moles HCl to produce 1 mol ZnCl2 and 1 mol H2

For 0.01071 moles HCl we'll have 0.005355 moles H2

Step 7: Calculate volume H2

1 mol at STP = 22.4 L

0.005355 moles = 22.4 * 0.005355 = 0.120 L = 120 mL

0.120 L of hydrogen gas will be produced

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Mashutka [201]

The value of X is 2

The total amount of sample taken is 2.00 g

The amount of sample left in the oven after drying is 1.565g

The amount of sample lost (mass of water driven out) = Total sample-Anhydrous salt left in the oven

                                 = 2.00 - 1.565

                                = 0.435 grams

The moles of anhydrous salt present in the hydrate = 1.565g/129.83g/mol = 0.01205

The moles of water present in the hydrate = 0.4350g/18.01g/mol = 0.02415

Therefore the ratio of these two are in 1:2 ratio

The complete chemical reaction is   CoCl2.2H20

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6 0
1 year ago
How many moles of hydrogen are in 5.2 moles of c7h18
alexandr1967 [171]
The percentage of hydrogen in C7H18 is calculated as follows:
[18/(12*7+1*8)]*100=18%
The amount of hydrogen in 5.2moles is given by:(18/100)*5.2=0.94moles
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4. Cuál es la aplicación de las reacciones de óxido-reducción en la vida diaria y en la industria?
12345 [234]

Answer:

<em>Dentro de las aplicaciones de la óxido-reducción se pueden encontrar:</em>

  1. <u><em>La obtención del aluminio a partir de la alúmina y la electrolisis.</em></u>
  2. <u><em>La obtención de cloro, hidrógeno e hidróxido de sodio a partir del cloruro de sodio y la electrolisis.</em></u>
  3. <u><em>La combustión interna de un motor a gasolina u otro combustible fósil.</em></u>
  4. <u><em>Las termoeléctricas, las cuales para generar energía realizan combustión de carbón.</em></u>
  5. <u><em>La galvanoplastia, donde para evitar la corrosión de un metal se recubre con otro metal más resistente, por ejemplo: el recubrimiento del acero con zinc.</em></u>
  6. <u><em>La pilas o baterías de las cuales se obtiene energía química</em></u><em>.</em>

Explanation:

<em>Como puedes ver en la respuesta, la óxido-reducción tiene diversas aplicaciones en la vida moderna, desde todos los tipos de combustión los cuales sirven para brindar energía o movilizarte, hasta todas las funciones que se le ha dado a la electrolisis y a la obtención de la energía por medios químicos, incluso se puede considerar una aplicación de la óxido-reducción la incorporación de antioxidantes en los alimentos, los cuales disminuyen la velocidad de descomposición de los mismos. </em>

3 0
3 years ago
En una estructura de concreto cuyo peso es de 8500 n se apoyo sobre un area de 25cm2,hallar la presion ejercida sobre su base
babymother [125]

Respuesta:

340 N/cm²

Explicación:

Paso 1: Información provista

Peso de la estructura (F): 8500 Newton

Area superficial (A): 25 cm²

Paso 2: Calcular la presión (P) ejercida por la estructura de concreto sobre su base

La presión es igual al cociente entre la fuerza ejercida y la superficie sobre la que se aplica.

P = F/A

P = 8500 N / 25 cm² = 340 N/cm²

5 0
3 years ago
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