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Nina [5.8K]
2 years ago
6

Using the equation 2H2 + O2 --> 2 H2O, if 192 g of O are produced, how many grams of hydrogen must react with it? Help me und

erstand how to do this?
Chemistry
1 answer:
Mamont248 [21]2 years ago
4 0
<span>To produce a molar conversion, you need to know the molar mass of each molecule. I presume you mean there are 192 grams of O2. The molecular weight of oxygen is 16 g/mol. Therefore, O2 is 32 g/mole.

   If there are 192 grams of O2, then:
 192 grams x (1 mole/32g) = 6 moles of O2.

   To react each mole of oxygen, you need 2 moles of hydrogen (H2). You can see this in the equation 2H2 + O2 --> 2 H20.
   To react 6 moles of O2, you need 12 moles of H2.
   Now that we have the total moles of hydrogen needed, we now use the molar mass of H2 (2 grams/mole)
   12 moles H2 x (2 grams/1 mol H2) = 24 grams of H2.</span>
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A mixture of 15.0 g of the anesthetic halothane (C2HBrClF3 197.4 g/mol) and 22.6 g of oxygen gas has a total pressure of 862 tor
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Answer : The partial pressure of C_2HBrClF_3 and O_2 are, 84 torr and 778 torr respectively.

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Molar mass of C_2HBrClF_3 = 197.4 g/mole

Molar mass of O_2 = 32 g/mole

First we have to calculate the moles of C_2HBrClF_3 and O_2.

\text{Moles of }C_2HBrClF_3=\frac{\text{Mass of }C_2HBrClF_3}{\text{Molar mass of }C_2HBrClF_3}=\frac{15.0g}{197.4g/mole}=0.0759mole

and,

\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{22.6g}{32g/mole}=0.706mole

Now we have to calculate the mole fraction of C_2HBrClF_3 and O_2.

\text{Mole fraction of }C_2HBrClF_3=\frac{\text{Moles of }C_2HBrClF_3}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.0759}{0.0759+0.706}=0.0971

and,

\text{Mole fraction of }O_2=\frac{\text{Moles of }O_2}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.706}{0.0759+0.706}=0.903

Now we have to partial pressure of C_2HBrClF_3 and O_2.

According to the Raoult's law,

p^o=X\times p_T

where,

p^o = partial pressure of gas

p_T = total pressure of gas

X = mole fraction of gas

p_{C_2HBrClF_3}=X_{C_2HBrClF_3}\times p_T

p_{C_2HBrClF_3}=0.0971\times 862torr=84torr

and,

p_{O_2}=X_{O_2}\times p_T

p_{O_2}=0.903\times 862torr=778torr

Therefore, the partial pressure of C_2HBrClF_3 and O_2 are, 84 torr and 778 torr respectively.

6 0
2 years ago
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