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asambeis [7]
3 years ago
15

When 1 mole of hydrogen gas (H2) reacts with excess oxygen to form water at a constant pressure, 241.8 KJ of energy is released

as heat. Calculate ΔH for a process in which 21.9 g sample of hydrogen gas (H2) reacts with excess oxygen at constant pressure.
Chemistry
1 answer:
kompoz [17]3 years ago
4 0

Answer:

\text{1540 kJ}

Explanation:

1. Gather all the information in one place

Mᵣ:   2.016  

          H₂ + ½O₂ ⟶ H₂O + 148.1 kJ

m/g:  21.9

2. Moles of H₂

n = \text{21.9 g H}_{2} \times \dfrac{\text{1 mol H}_{2}}{\text{2.016 g H}_{2}} = \text{10.86 mol H}_{2}

3. Calculate ΔH

Treat the heat AS IF it were a reactant or product. In this case, the reaction is exothermic, so the heat is a product.

In effect, you have 148.1 mol of "kJ"s for each mole of hydrogen and

\Delta \text{H} = \text{10.86 mol H}_{2} \times \dfrac{\text{141.8 kJ}}{\text{1 mol H}_{2}} = \textbf{1540 kJ}

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What are the three types of plate boundaries? What is the direction of movement at each boundary?
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8 0
3 years ago
How many grams of H2O will be formed when 32.0 g H2 is mixed with 84.0 g of O2 and allowed to react to form water
Zarrin [17]

Answer:

94.58 g of H_2O

Explanation:

For this question we have to start with the reaction:

H_2~+~O_2~->~H_2O

Now, we can balance the reaction:

2H_2~+~O_2~->~2H_2O

We have the amount of H_2  and the amount of O_2 . Therefore we have to find the limiting reactive, for this, we have to follow a few steps.

1) Find the moles of each reactive, using the molar mass of each compound (H_2~=~2~g/mol~~O_2=~32~g/mol ).

2) Divide by the coefficient of each compound in the balanced reaction ("2" for H_2 and "1" for O_2).

<u>Find the moles of each reactive</u>

32.0~g~H_2\frac{1~mol~H_2}{2~g~of~H_2}=15.87~mol~H_2

84.0~g~of~O_2\frac{1~mol~of~O_2}{32~g~of~O_2}=2.62~mol~of~O_2

<u>Divide by the coefficient</u>

<u />

\frac{15.87~mol~H_2}{2}=7.94

\frac{2.62~mol~of~O_2}{1}=2.62

The smallest values are for H_2, so hydrogen is the limiting reagent. Now, we can do the calculation for the amount of water:

32.0~g~H_2\frac{1~mol~H_2}{2~g~of~H_2}\frac{2~mol~H_2O}{2~mol~H_2}\frac{18~g~H_2O}{1~mol~H_2O}=94.58~g~H_2O

We have to remember that the molar ratio between H_2O and H_2 is 2:2 and the molar mass of H_2O is 18 g/mol.

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