3. ¿Cuál de las siguientes operaciones es correcta para calcular el número de moles de hidrógeno necesarios para producir 6 mole
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<span>Molar mass (MM) of benzene C6H6
C = 6 * 12 = 72u
H = 6 * 1 = 6u
MM C6H6 = 72 + 6 = 78 g / mol
Benzene - Molar Mass = 78 g --------- 1 mol
Of A Mix has 468 g -------------- x
78x = 468
X = 468/78
X = 6 moles
Molar mass (MM) of Hydrochloric Acid HCl H = 1 * 1 = 1u
CI = 1 * 35 = 35u
MM HCl = 1 + 35 = 36 g / mol
Hydrochloric Acid - Molar Mass = 36 g ---------- 1 mol
Of A Mix has 72 g ------------ y
36y = 72
Y = 72/36
Y = 2 moles
Thus, a mixture has a total of 8 moles (6 mol + 2 mol).
Dividing One Mole Amount of Each Substance by the Number of Total Mole Amounts, Then we will obtain a Molar Fraction of Each:
Molar fraction make benzene = (6/8) simplify 2 = 3/4
Molar Fraction to make Hydrochloric Acid = (2/8) = simplify 2 = 1/4
Note:. The sum of the molar fractions of the always give goes 1, we have: 3/4 + 1/4 = 1
ANSWER:
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C = 6 * 12 = 72u
H = 6 * 1 = 6u
MM C6H6 = 72 + 6 = 78 g / mol
Benzene - Molar Mass = 78 g --------- 1 mol
Of A Mix has 468 g -------------- x
78x = 468
X = 468/78
X = 6 moles
Molar mass (MM) of Hydrochloric Acid HCl H = 1 * 1 = 1u
CI = 1 * 35 = 35u
MM HCl = 1 + 35 = 36 g / mol
Hydrochloric Acid - Molar Mass = 36 g ---------- 1 mol
Of A Mix has 72 g ------------ y
36y = 72
Y = 72/36
Y = 2 moles
Thus, a mixture has a total of 8 moles (6 mol + 2 mol).
Dividing One Mole Amount of Each Substance by the Number of Total Mole Amounts, Then we will obtain a Molar Fraction of Each:
Molar fraction make benzene = (6/8) simplify 2 = 3/4
Molar Fraction to make Hydrochloric Acid = (2/8) = simplify 2 = 1/4
Note:. The sum of the molar fractions of the always give goes 1, we have: 3/4 + 1/4 = 1
ANSWER:
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