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2 years ago

How many grams of iron (Fe) would be needed to generate 5.25 g of hydrogen gas (H2)?

Chemistry

1 answer:

Free_Kalibri [48]2 years ago

8 0

Considering the given reaction, 109 g of Fe are needed to generate 5.25 g of H₂.

<h3>What is stoichiometry?</h3>

Stoichiometry refers to the relationship between the quantities of reactants and products before, during, and following chemical reactions.

Step 1: Write the balanced equation.

3 Fe + 4 H₂O ⇒ 4 H₂ + Fe₃O₄

Step 2: Calculate the mass of Fe needed to produce 5.25 g of H₂.

We will consider the following relationships.

The molar mass of Fe is 55.85 g/mol.
The molar ratio of Fe to H₂ is 3:4.
The molar mass of H₂ is 2.02 g/mol.

$5.25gH_2 \times \frac{1molH_2}{2.02gH_2} \times\frac{3molFe}{4molH_2} \times\frac{55.85gFe}{1molFe} = 109gFe$

Considering the given reaction, 109 g of Fe are needed to generate 5.25 g of H₂.

Learn more about stoichiometry here: brainly.com/question/16060223

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PLEASE HELP & SHOW WORK

Olegator [25]

Answer:

1. 0.178 moles ; 2. 8x10²³ atoms ; 3. 7.22x10²³ molecules ; 4. 89.6 g ; 5. 1.34x10²² atoms ; 6. 1.67x10²⁵ molecules

Explanation:

1. Mass / Molar mass = Mol

5g / 28 g/m = 0.178 moles

2. 1 molecule of N₂ has 2 atoms, it is a dyatomic molecule.

4x10²³ x2 = 8x10²³ atoms

3. 1 mol of anything, has 6.02x10²³ particles

6.02x10²³ molecules . 1.2 mol = 7.22x10²³

4. 1 atom of C weighs 12 amu.

4.5x10²⁴ weigh ( 4.5x10²⁴ . 12) = 5.24x10²⁵ amu

1 amu = 1.66054x10⁻²⁴g

5.24x10²⁵ amu = (5.24x10²⁵ . 1.66054x10⁻²⁴) = 89.6 g

5. Molar mass NaCl = 58.45 g/m

1.3 g / 58.45 g/m = 0.0222 moles

1 mol has 6.02x10²³ atoms

0.0222 moles → ( 0.0222 . 6.02x10²³) = 1.34x10²²

6. Density of water is 1 g/mL, so 500 mL are contained in 500 g of water

Molar mass H₂O = 18 g/m

500 g / 18 g/m = 27.8 moles

6.02x10²³ molecules . 27.8 moles = 1.67x10²⁵

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3 years ago

How much heat must your body transfer to 500.0g of water to heat it from 25.0°C to body temperature, 37.0°C?

shtirl [24]

Answer : The heat your body transfer must be, 25.1 kJ

Explanation :

Formula used :

$Q=m\times c\times \Delta T$

or,

$Q=m\times c\times (T_2-T_1)$

where,

Q = heat = ?

m = mass of water = 500.0 g

c = specific heat of water = $4.18J/g^oC$

$T_1$ = initial temperature = $25.0^oC$

$T_2$ = final temperature = $37.0^oC$

Now put all the given value in the above formula, we get:

$Q=500.0g\times 4.18J/g^oC\times (37.0-25.0)K$

$Q=25080J=25.1kJ$

Therefore, the heat your body transfer must be, 25.1 kJ

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