Molar Volume Calculations
暂无描述。系统推荐的高质量记忆内容，适合每天坚持背诵学习。
卡片预览 (10 张)
what is a molar volume
a volume of gas that contains one mole of that gas
how do the molar volumes vary for all gases for given temperatures and pressures
they are approximately the same
what are the most often used values for the given temperature and pressure and what is it abbreviated to
• the temperature is room temp at 298K • while the pressure is at the standard pressure of 1atm (101325 kPa) • the abbreviation is RTP (room temperature pressure)
what is the molar volume for all gases at RTP and what is its symbol
• 24dm3 or 24000cm3 - Vm = 24dm3mol-1
using the units of Vm, what would the equation for molar volume be
molar volume = volume of gas (dm or cm) / amount in moles
what is the amount in moles of CO in 3.8dm3 of cabon monoxide
• one mole of CO at RTP would be 24cm3 • so to work out how many moles you have rearrange the equation • giving moles = volume of gas / molar volume • 3.8 / 24 = 0.158 mols
what is the volume of 0.356mol of hydrogen gas
• rearrange to volume of gas = moles x molar volume - 24 x 0.356 = 8.54dm3
a piece of Mg with mass 1g is added to an excess of dilute hydrochloric acid. What volume of hydrogen gas is formed? The equation is Mg + 2HCl = MgCl2 + H2
• you need to work out the moles of Mg there are • you can then use this to compare the ratio / coefficients in the equation for how many moles of hydrogen gas there would be • as one equation mole of Mg produces one e-mole of hydrogen gas, the moles of Mg used would be equal to hydrogens • moles = mass / Mr, so 1 / 24/3 = 0.0412 mols • volume of gas = moles x molar volume, so 0.0412 x 24 = 0.988dm3
in the reaction CaCO3 + 2NHO3 = Ca(NO3)2 + H2O + CO2, 100cm3 of carbon dioxide is formed. What mass of calcium carbonate is needed for this?
• work out the moles of carbon dioxide produced • the ratio of the coefficients between calcium carbonate and carbon dioxide is 1:1 so you need the same number of moles of CaCO3 • moles = volume / molar mass so 100 / 24000 = 0.004167 • mass = moles x Mr, so 100 x 0.004167 = 0.417g
in the reaction (NH4)2SO4 + 2NaOH = Na2SO4 + 2H2O + 2NH3, what volume of ammonia is formed by reacting 2.16g of ammonium sulfate with excess sodium hydroxide
• the same steps as the others, but this time the ratio coefficients are 1:2 • Mr of (NH4)2SO4 = 132gmol-1 • moles = mass / Mr so 2.16 / 132 = 0.0164 mols • as one e-mole of that makes 2 e-moles of ammonia, you times it by 2 to get 0.0328 • then you do volume = moles x molar volume to get 0.0328 x 24 = 0.785dm3