<h2>Answer:</h2>
Moles of a gas = 0.500
Volume = 2.50 L
Pressure = 13. atm
Temperature = ?
Solution:
Formula:
PV = n RT
Putting the values in formula:
T = PV/nR = 13 * 2.5 / 0.5 * 0.082057
= 32.5/0.041 = 792.68 K
T = 792.68 K
Answer: D. 19.9 g hydrogen remains.
Explanation:
To calculate the moles, we use the equation:
a) moles of 
b) moles of 
According to stoichiometry :
1 mole of require 1 mole of
Thus 0.0787 moles of 
require=
of
Thus is the limiting reagent as it limits the formation of product and acts as the excess reagent. (10.0-0.0787)= 9.92 moles of are left unreacted.
Mass of 
Thus 19.9 g of remains unreacted.
Answer:
204.5505 grams
2.5666 moles
Explanation:
For the first question, multiply 3.5 (# of moles) by 58.443 (g/mol for NaCl).
58.443 * 3.5
<em>I'll distribute 3.5 into 58.443.</em>
(3.5 * 50) + (3.5 * 8) + (3.5 * 0.4) + (3.5 * 0.04) + (3.5 * 0.003)
175 + 28 + 1.4 + 0.14 + 0.0105
203 + 1.4 + 0.14 + 0.0105
204.4 + 0.14 + 0.0105
204.54 + 0.0105
204.5505 grams
There are 204.5505 grams in 3.5 moles of NaCl.
For the second question, divide 150 (# of grams) by 58.443 (g/mol for NaCl). I'll convert both into fractions.
150/1 * 1000/58443
150000/58443
2.56660336 moles
2.5666 moles (rounded to 4 places to keep consistency with the first answer) are in 150 grams of NaCl.
