10. What mass of LiSO4 (110 g/mol) is required to prepare 30mL of 0.20M LiSO4?
11. Liquid bromine is reacted with aluminum to produce solid aluminum bromide. If 0.450L of bromine reacts with excess aluminum, how much aluminum is made? (Density of liquid bromine is 3.1g/L)
I worked the Li2SO4 problem below.
Sure! Let's break down each question and explain how to find the answers step by step:
Question 10: What mass of LiSO4 (110 g/mol) is required to prepare 30 mL of 0.20 M LiSO4?
To find the answer, we need to use the formula:
mass = moles * molar mass
First, we need to calculate the number of moles of LiSO4 required:
moles = volume * concentration
Given:
volume = 30 mL
concentration = 0.20 M
Converting the volume from mL to L:
volume = 30 mL * (1 L / 1000 mL) = 0.03 L
Now, substituting the values into the formula:
moles = 0.03 L * 0.20 M = 0.006 moles
Next, we calculate the mass using the molar mass of LiSO4:
mass = 0.006 moles * 110 g/mol = 0.66 g
Therefore, 0.66 grams of LiSO4 is required to prepare 30 mL of 0.20 M LiSO4.
Question 11: Liquid bromine is reacted with aluminum to produce solid aluminum bromide. If 0.450 L of bromine reacts with excess aluminum, how much aluminum is made? (Density of liquid bromine is 3.1 g/L)
To find the answer, we need to calculate the mass of bromine using its density, and then use stoichiometry to determine the mass of aluminum used.
First, calculate the mass of bromine using its density:
mass = volume * density
Given:
volume = 0.450 L
density = 3.1 g/L
mass = 0.450 L * 3.1 g/L = 1.395 g
Now, we can use the stoichiometry of the balanced chemical equation to determine the mass of aluminum produced. As the reaction is given as a ratio, we can use the balanced equation as follows:
1 mole of bromine reacts with 2 moles of aluminum
To find the moles of aluminum used, we need to convert the mass of bromine to moles:
moles of bromine = mass of bromine / molar mass of bromine
Given:
molar mass of bromine = 79.904 g/mol
moles of bromine = 1.395 g / 79.904 g/mol = 0.0175 moles
Since the ratio of bromine to aluminum is 1:2, we can use the stoichiometry to calculate the moles of aluminum:
moles of aluminum = 2 * moles of bromine = 2 * 0.0175 moles = 0.0350 moles
Finally, convert the moles of aluminum to mass using its molar mass:
mass of aluminum = moles of aluminum * molar mass of aluminum
Given:
molar mass of aluminum = 26.982 g/mol
mass of aluminum = 0.0350 moles * 26.982 g/mol = 0.944 g
Therefore, approximately 0.944 grams of aluminum is made when 0.450 L of bromine reacts with excess aluminum.