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kykrilka [37]
2 years ago
15

What mass of CCl4 is required to prepare a 0.25 m solution using 115 g of hexane? (molar mass of ccl4 = 153.81 g/mol and molar m

ass of hexane = 86.17 g/mol.)
Chemistry
1 answer:
natka813 [3]2 years ago
7 0

4.42 g mass of CCl4 is required to prepare a 0.25 m solution using 115 g of hexane.

It's easy to find the molecular mass of a compound with these steps: Determine the molecular formula of the molecule. Use the periodic table to determine the atomic mass of each element in the molecule. Multiply each element's atomic mass by the number of atoms of that element in the molecule.

The molar mass of any compound can be found out by adding the relative atomic masses of each element present in that particular compound.

Hexane is an organic compound, a straight-chain alkane with six carbon atoms and has the molecular formula C₆H₁₄.

Therefore,

⇒ 0.115 g of Hexane x (0.25 mol CCl4/1 mol hexane) x (153.81 g of CCl4/1 mol CCl4) = 4.42g CCl4.

To learn more about CCL4 and Hexane here

brainly.com/question/15156642

#SPJ4

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Hydrogen gas and oxygen gas react to form liquid water according to the following equation:

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a. Converting our given masses of each gas to moles, we have:

(25 g H2)/(2 × 1.008 g/mol) = 12.4 mol H2; and

(25 g O2)/(2 × 15.999 g/mol) = 0.781 mol O2.

From the equation, two moles of H2 react with every one mole of O2. To fully react with 12.4 moles of H2, as we have here, one would need 6.2 moles of O2, which is far more than what we're actually given. Thus, the oxygen is our limiting reactant, and as such it will be the first reactant to run out.

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