Five grams of sodium are combined with unlimited bromine. How many moles of bromine are used up? How many liters of bromine would be used if the reaction took place at STP?
My answer:
.1088 mole Br2 and 2.44 L Br2
If that is 5 g then you have only one significant figures so the answer is 0.1 and you're correct. However, the second part can't be done that way. You have assume, I suppose, that Br2 is a gas and it occupies 22.4 L/mol. If that were true your answer of 2.44 would be ok BUT Br2 is not a gas at STP. I think the author of the problem screwed up. If you look up the density of liquid Br2, then mass = volume x density. With the density and the mass you could calculate the volume.
To find the number of moles of bromine used up, we first need to determine the number of moles of sodium using its molar mass. Sodium has a molar mass of 22.99 g/mol.
Given that the mass of sodium is 5 grams, we can calculate the number of moles of sodium:
moles of sodium = mass of sodium / molar mass of sodium
= 5 g / 22.99 g/mol
≈ 0.2175 mol Na
From the balanced chemical equation, we know that the ratio of sodium to bromine is 2:1. That means for every 2 moles of Na, 1 mole of Br2 is used up.
So, the number of moles of bromine used up can be calculated as follows:
moles of bromine used up = moles of sodium / 2
≈ 0.2175 mol Na / 2
≈ 0.1088 mol Br2
Now, to find the volume of bromine used at STP, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (at STP, it is 1 atm)
V = volume
n = number of moles of bromine
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature (at STP, it is 273.15 K)
Rearranging the equation, we get:
V = nRT / P
Now, we can substitute the values into the equation:
V = (0.1088 mol) * (0.0821 L.atm/mol.K) * (273.15 K) / (1 atm)
≈ 2.44 L Br2
So, the number of moles of bromine used up is approximately 0.1088 mol Br2, and the volume of bromine used at STP is approximately 2.44 L Br2.