The important thing in this question is the unit. The mass equals density * volume. 3.1 L = 3.1 * 10^3 cm3. So the mass is 3.193*10^3 g. 1 pound = 453.95 g. So the answer is 7.04 pounds.
The balanced reaction would be:
2CO + O2 = 2CO2
We assume that the gases are ideal gas so that we use the relation that 1 mol of an ideal gas is equal to 22.4 L of the gas at STP. From that relation, we get the number of moles and we can convert it to other units. We do as follows:
1.0 L CO ( 1 mol / 22.4 L ) ( 2 mol CO2 / 2mol CO ) = 0.045 mol CO2 produced
0.045 mol CO2 ( 22.4 L / 1 mol ) = 1 L of CO2
0.045 mol CO2 ( 44.01 g / 1 mol ) = 1.98 g of CO2
Answer:
Percentage lithium by mass in Lithium carbonate sample = 19.0%
Explanation:
Atomic mass of lithium = 7.0 g; atomic mass of Chlorine = 35.5 g; atomic mass of carbon = 12.0 g; atomic mass of oxygen = 16.0 g
Molar mass of lithium chloride, LiCl = 7 + 35.5 = 42.5 g
Percentage by mass of lithium in LiCl = (7/42.5) * 100% = 16.4 % aproximately 16%
Molar mass of lithium carbonate, Li₂CO₃ = 7 * 2 + 12 + 16 * 3 =74.0 g
Percentage by mass of lithium in Li₂CO₃ = (14/74) * 100% = 18.9 % approximately 19%
Mass of Lithium carbonate sample = 2 * 42.5 = 85.0 g
mass of lithium in 85.0 g Li₂CO₃ = 19% * 85.0 g = 16.15 g
Percentage by mass of lithium in 85.0 g Li₂CO₃ = (16.15/85.0) * 100 % = 19.0%
Percentage lithium by mass in Lithium carbonate sample = 19.0%