Answer:
1. d. 84g
2. c. 77.4%
Explanation:
1. Based on the reaction 2 moles of Na produce 2 moles of NaCl.
The moles of Na (22.99g/mol) in 33g are:
33g * (1mol / 22.99g) = 1.435 moles Na = 1.435 moles NaCl must be produced.
The mass of NaCl is (Molar mass: 58.44g/mol):
1.435moles * (58.44g / mol) = 84g
d. 84g
2. Percent yield is defined as 100 times the ratio between actual yield (65g) and theoretical yield (84g). The percent yield is:
65g / 84g * 100 =
c. 77.4%
Correct formula for what?
Answer:
The volume of sample at STP is = 41.22 ml
Explanation:
Given data
= 0.9 atm
= 50 ml
= 298 K
At STP
= 1 atm
= 273 K
From ideal gas equation
= 41.22 ml
Therefore the volume of sample at STP is = 41.22 ml
<span>3.68 liters
First, determine the number of moles of butane you have. Start with the atomic weights of the involved elements:
Atomic weight carbon = 12.0107
Atomic weight hydrogen = 1.00794
Atomic weight oxygen = 15.999
Molar mass butane = 4*12.0107 + 10*1.00794 = 58.1222 g/mol
Moles butane = 2.20 g / 58.1222 g/mol = 0.037851286
Looking at the balanced equation for the reaction which is
2 C4H10(g)+13 O2(g)→8 CO2(g)+10 H2O(l)
It indicates that for every 2 moles of butane used, 8 moles of carbon dioxide is produced. Simplified, for each mole of butane, 4 moles of CO2 are produced. So let's calculate how many moles of CO2 we have:
0.037851286 mol * 4 = 0.151405143 mol
The ideal gas law is
PV = nRT
where
P = Pressure
V = Volume
n = number of moles
R = Ideal gas constant ( 0.082057338 L*atm/(K*mol) )
T = absolute temperature (23C + 273.15K = 296.15K)
So let's solve the formula for V and the calculate using known values:
PV = nRT
V = nRT/P
V = (0.151405143 mol) (0.082057338 L*atm/(K*mol))(296.15K)/(1 atm)
V = (3.679338871 L*atm)/(1 atm)
V = 3.679338871 L
So the volume of CO2 produced will occupy 3.68 liters.</span>
