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Nutka1998 [239]
3 years ago
15

Molten gallium reacts with arsenic to form the semiconductor, gallium arsenide, GaAs, used in light-emitting diodes and solar ce

lls: Ga(I) +As(s) GaAs(s)
a. If 4.00 g of gallium is reacted with 5.50 g of arsenic, how many grams of the excess reactant are left at the end of the reaction?
b. If 4.00 g of gallium is reacted with 4.94 g of arsenic, how many grams of the excess reactant are left at the end of the reaction?
c. If 4.00 g of gallium is reacted with 0.56 g of arsenic, how many grams of the excess reactant are left at the end of the reaction?
d. If 8.94 g of gallium is reacted with 5.50 g of arsenic, how many grams of the excess reactant are left at the end of the reaction?
e. If 4.00 g of gallium is reacted with 1.50 g of arsenic, how many grams of the excess reactant are left at the end of the reaction?
Chemistry
1 answer:
Vsevolod [243]3 years ago
7 0

Answer:

a) 1.2g of arsenic b) 0.64g of arsenic c) 3.481g of gallium d) 3.806g of gallium e) 2.61g arsenic

Explanation:

The balanced equation is:

Ga + As = GaAs, 1:1 mole ratio

a) mass (gallium)/ molar mass of Ga = 4/ 69.723 = 0.0574mol

Mass (arsenic)/ molar mass of As = 5.5/74.9216 = 0.0734, subtracting the moles from each other (knowing already that the ratio is 1:1), arsenic is in excess by 1.2g

b) repeating the procedure (changing the values)

It will be 0.0574 to 0.06593

Arsenic is in excess by 0.00854

0.00854* mass of arsenic (such must be done for the first remaining mole) = 0.64g

c) the mole ratio is 0.0574: 0.00747

Gallium is in excess by 0.05

Mass of excess gallium = 0.05* 69.723 = 3.481g

d) using the mass given, the new ratio is

0.128: 0.0734

Gallium is in excess by 0.054mol

Mass of excess gallium = 0.054*69.723 = 3.806g

e) using the mass again, the new ratio is 0.0574: 0.02,

Gallium is in excess by 0.0374*69.723 = 2.61g

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Answer:

1. 9.4 grams of methane produce<u> 25.85</u> grams of CO2

2.Grams of water produced = <u>11.81 grams</u>

3.Mass of Methane produced by 10.1 gram of O2 = <u>2.52 grams</u>

4.Amount of methane consumed = <u>46.9 grams</u>

5. Grams of Co2 produced =<u> 8.32 grams</u>

<u></u>

Explanation:

Molar masses :

Methane = CH4 = mass of C + 4x (mass of H)

CH4 = 12 +4(1) = 16 grams

<u>1 mole of CH4 = 16 gram</u>

Oxygen O2 = 2 x (mass of O) = 2x(16) = 32 gram (1 mole of O2 =32 gram)

Carbon Dioxide =CO2 = mass of C + 2(mass of O)

= 12 + 2(16)

= 44 grams <u>(1 mole of CO2 = 44 gram )</u>

Water = H2O = 18 grams ( 1 mole of H2O = 18 gram)

1 mole of each molecule is equal to their molar masses

The balanced equation is :

1CH_{4}(g)+2O_{2}\rightarrow 1CO_{2}+2H_{2}O(l)

According to Stoichiometry :

1 mole of CH4 = 2 Mole of O2 = 1 mole of CO2 = 2 mole of H2O

1. From the equation ,

1 mole of methane produce  =1 mole of CO2

16 gram of methane = 44 gram of CO2

1 gram of methane =

\frac{44}{16} gram of CO2

9.4 gram of CH4 =

\frac{44}{16}\times 9.4 gram of CO2

= 25 .85 gram of CO2

2.

2 mole of O2 produces = 2 mole of H2O(water)

1 mole of O2 produces = 1 mole of H2O

32 gram of O2 = 18 gram of water

1 gram of O2 =

\frac{18}{32}

21 gram of O2 =

\frac{18}{32}\times 21

11.81 gram of water

3. 1 mole of CH4 = 2 mole of O2

16 gram of CH4 = 2(32)  = 64 grams of O2

64 gram of O2 needs = 16 grams of CH4

1 gram of O2 needs =

\frac{16}{64}

10.1 gram of O =

\frac{16}{64}\times 10.1 of CH4

= 2.52 gram

4.

1 mole of CO2 is produced from = 1 mole of CH4

44 gram of CO2 is produced from 16 gram of CH4

1 gram CO2 =

\frac{16}{44} gram of CH4

129 gram of CO2 =

\frac{16}{44}\times 129 gram of CH4

= 46.90 grams

5.

2 mole of O2  produce = 1 mole of CO2

2x 32 gram of O2 = 44 gram of CO2

1 gram of O2 =

\frac{44}{64} of CO2

12.1 gram of O2 produce=

\frac{44}{64}\times 12.1 of CO2

= 8.318 gram

Note : Write the quantity give on left side of "="

write the substance asked on right side of "="

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The mass of hydrazine (N₂H₄) required to produce 96 g of water (H₂O) is 85.4 g (Option C)

<h3>Balanced equation </h3>

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Molar mass of N₂H₄ = (2×14) + (4×1) = 32 g/mol

Mass of N₂H₄ from the balanced equation = 1 × 32 = 32 g

Molar mass of H₂O = (2×1) + 16 = 18 g/mol

Mass of H₂O from the balanced equation = 2 × 18 = 36 g

SUMMARY

From the balanced equation above,

36 g of H₂O were produced by 32 g of N₂H₄

<h3>How to determine the mass of N₂H₄</h3>

From the balanced equation above,

36 g of H₂O were produced by 32 g of N₂H₄

Therefore,

96 g of H₂O will be produced by = (96 × 32) / 36 = 85.4 g of N₂H₄

Thus, 85.4 g of N₂H₄ is needed for the reaction

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