Most reactive to least reactive out of Cu, Ca, Pb, Sn, Mg, K, and Na.

How many grams of ethane gas (C2H6) are in 12.7-liter sample at 1.6 atmosphere and 24 degrees Celsius? Show all work used to sol

pentagon [3]

Data:
M (Molar Mass) of C2H6
C = 2*12 = 24 amu
H = 6*1 = 6 amu
---------------------
MM C2H6 = 24+6 = 30 g/mol

P (pressure) = 1.6 atm
V (volume) = 12.7 L
R = 0,082 atm .L/mol.K
n (Number of mols) → $n = \frac{m}{MM} m (mass) = ? T = 24ºC Celsius to Kelvin TK = TºC + 273 TK = 24 + 273 TK = 297 By the equation of state of the gases or equation of Clapeyron, we have:
 [tex]P*V = n*R*T$ 

Since $n = \frac{m}{MM}$ , we can perform the following substitution in the above Clapeyron equation:


$P * V = n * R * T$
$P*V = \frac{m}{MM} *R*T$
multiply cross
$m = \frac{P*V*MM}{R*T}$

Solving:
$m = \frac{P*V*MM}{R*T}$
$m = \frac{1.6*12.7*30}{0.082*297}$
$m = \frac{609.6}{24.354}$
$\boxed{\boxed{m \approx 25.03\:grams}}\end{array}}\qquad\quad\checkmark$

Answer:
25.03 grams of ethane gas

5 0

3 years ago

Consider the above reaction using 25.0 mL of tert-butyl alcohol (d = 0.786 g/mL) with 60.0 mL of concentrated hydrobromic acid (

Liono4ka [1.6K]

Answer:

Theoretical yield: 36,3g

Percent yield: 71,9%

Explanation:

The reaction of tert-butyl alcohol with hydrobromic acid to produce tert-butyl bromide is a 1:1 reaction.

Moles of tert-butyl alcohol in 25,0 mL are:

25,0mL × (0,786g / mL) × (1mol / 74,12g) = 0,265 moles of tert-butyl alcohol

Moles of hydrobromic acid are: