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11Alexandr11 [23.1K]
3 years ago
12

What mass of H2 is needed to react with 8.75 g of O2 according to the following equation: O2(g) + H2(g) → H2O(g)?

Chemistry
2 answers:
Alika [10]3 years ago
7 0

Answer:- B: 1.10g H_2 is the right answer.

Solution:- The balanced equation is:

O_2(g)+2H_2(g)\rightarrow 2H_2O(g)

We have been given with 8.75 grams of oxygen and asked to calculate the grams of hydrogen needed to react with given grams of oxygen according to the balanced equation.

From balanced equation, 1 mole of oxygen reacts with 2 moles of hydrogen.

So, let's convert grams of oxygen to moles and multiply it by the mole ratio to calculate the moles of hydrogen that are easily converted to grams on multiplying by it's molar mass.

The complete set up looks as:

8.75g O_2(\frac{1mole}{32g})(\frac{2mole H_2}{1mole O_2})(\frac{2.02g}{1mole})

= 1.10g H_2

Hence, the right option is B: 1.10g H_2 .



horrorfan [7]3 years ago
7 0

Answer:

1.10 g H2

Explanation:

I took the test and got it right.

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C3H8 + 5O2 → 3CO2+ 4H2O, if 5.75L of oxygen are consumed in the above reaction, how many L of carbon dioxide are produced?
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Answer: 3.45 L carbon dioxide are produced

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number 6.023\times 10^{23} of particles.

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given volume}}{\text {Molar volume}}=\frac{5.75L}{22.4L}=0.257moles

C_3H_8+5O_2(g)\rightarrow 3CO_2+4H_2O  

According to stoichiometry :

5 moles of O_2 produce =  3 moles of CO_2

Thus 0.257 moles of O_2 will produce=\frac{3}{5}\times 0.257=0.154moles  of CO_2  

Volume of CO_2=moles\times {\text {Molar volume}}=0.154moles\times 22.4L/mol=3.45

Thus 3.45 L carbon dioxide are produced

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