what mass of O2 will react with 24 g of Mg to produce MgO,,
first and most important, is you need a balanced reaction. can you do that?
yup
\[2Mg + O2\rightarrow 2MgO\]
options are 32,16,24,8
good. so, you know the coefficients in front of the formulas tell us a MOLAR ratio, not a MASS ratio, right?
so is it 32
why do you think it's 32?
yup so do i need limiting reactant calculations
this isn't a limiting reactant problem, since it's the mass of oxygen you're looking for. this is simpler than that
well O2 has 32 mass and on right side it's 2MgO meands
*means that whole O2 is used right???
but you're told 24g of Mg, not MgO. What's special about 24g of Mg? Like something you'd find on the periodic table
i'm only concerned here about the mass of O2 which reacts in this reaction
but to know that, you need to know about the Mg, so what's special about 24g of Mg?
it's the mass of one mole of Mg
exactly
so you have 1 mole of Mg that will be used. What's the ratio of Mg to O2 in the balanced reaction?
2:1
so every time 2 atoms (or moles) of Mg react, only 1 O2 will react.
ohh so it's 16 g of O2 reacting cuz i'm asked to find the mass reacting with 1 mole of Mg and
So from that, if you use 1 mole of Mg, you'll only use ..... HALF a mole of O2
means 16g right????
yes
thanks
can u help me over one more question
@JFraser
sure
1 dm cube of an unknown gas weighs 1.25 g at STP. the possible formula for the gas is CO2, CO, C2H2 , SO2
you have a volume, a temp, and a pressure. you need to find the molar mass using the ideal gas law.
\[PV = nRT\]so\[n = \frac{PV}{RT}\]find the moles of gas, and divide the mass you're given by the moles that you calculate
yup i have 1atm pressure 1K temp and 1dm cube of volume
temp isn't 1K, STP is 273K
ohh sorry got confused
but i'm not given any mass
yes you are, read the problem
ohh yup and after doing this how can i predict the molecular formula of the gas
each of the gases will have its own molar mass. find the molar mass of each option, and the one that is right is the one that is the closest to your value
thanx again
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