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max2010maxim [7]
3 years ago
13

A student performed a chemical reaction in which 35 grams of hydrogen and 65 grams of oxygen reacted to form water. What is the

mass of the products? (answer with a number)
Chemistry
1 answer:
sergejj [24]3 years ago
8 0

Answer:

The mass of the product is 73,06 g

Explanation:

The reaction to form water is:

2H2 + O2 ---> 2H2O

35 g     65 g

Molar mass H2 = 2g/m ---> I have 17,5 moles (35g / 2 g/m)

Molar mass O2 = 32g/m ---> I have 2,03 moles (65g / 32g/m)

2 moles of H2 __reacts with __ 1 mol O2

17,5 moles of H2 __ reacts with ___ (17,5 m . 1 m) / 2 m = 8,75 m

I have just 2,03 m of O2 and I need 8,75 m so O2 is my limiting reagent

1 mol of O2 ___ reacts with 2 moles H2

2,03 moles of O2 __ reacts with  (2,03 m . 2 m) / 1 m = 4,06 m

I have 17,5 moles of H2 and I need 4,06 m. H2 is my excess reagent

<u><em>"All operations are done with the limiting reagent"</em></u>

1 mol O2 are required____ to form 2 H2O

2,03 mol O2 are required __ to form (2,03m . 2m) / 1m = 4,06 m

Molar mass of water: 18 g/m

Mass of water ----> Molar mass . Moles water = 18 g/m . 4,06 m = 73,08g

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Answer:

As compared to metals, they have low density and will melt at low temperatures. The shape of nonmetals cannot be changed easily because they are brittle and will break. Elements that have properties of both metals and nonmetals are called metalloids. They can be shiny or dull and their shape is easily changed.

Explanation:

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The answer is the first option (a)
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Name the following bases<br> -Ba(OH)2<br> - Ca(OH)2<br> -RbOH
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Ba(OH)_2\rightarrow \bold{\green{Barium\: hydroxide}}

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What mass of iron is formed when 10 grams of carbon react with 80 grams of iron iii oxide?
yKpoI14uk [10]

Answer:

55.85 grams of Fe is formed.

Explanation:

Identify the reaction:

2Fe₂O₃  +  3C  →  4Fe  +  3CO₂

Identify the limiting reactant, previously determine the mol of each reactant

(mass / molar mass)

10 g / 12 g/m = 0.83 moles C

80 g / 159.7 g /m = 0.500 moles Fe₂O₃

2 moles of oxide need 3 moles of C, to react

0.5 moles of oxide, will need ( 0.5  . 3)/ 2 = 0.751 mol

I have 0.83 moles of C, so C is the excess.

The limiting is the oxide.

3 mol of C need 2 mol of oxide to react

0.83 mol of C, will need (0.83  . 2)/ 3 = 0.553 mol of oxide, and I only have 0.5 (That's why Fe₂O₃ is the limiting)

Ratio is 2:4 (double)

If I have 0.5 moles of oxide, I will produce the double, in the reaction.

1 mol of Fe, will be produce so its mass is 55.85 g

5 0
3 years ago
A sample of water is heated from 60.0 °C to 75.0°C by the addition of 140 j of
kupik [55]

Mass of the water : 2.23 g

<h3>Furter explanation</h3>

Heat

Q = m.c.Δt

m= mass, g

c = heat capacity, for water : 4.18 J/g° C.

ΔT = temperature

Q= 140 J

Δt = 75 - 60 = 15

mass of the water :

\tt m=\dfrac{Q}{c.\Delta T}=\dfrac{140}{4.18\times 15}=2.23~g

5 0
3 years ago
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