<span>this is a limiting reagent problem.
first, balance the equation
4Na+ O2 ---> 2Na2O
use both the mass of Na and mass of O2 to figure out how much possible Na2O you could make.
start with Na and go to grams of Na2O
55.3 gNa x (1molNa/23.0gNa) x (2 molNa2O/4 molNa) x (62.0gNa2O/1molNa2O) = 75.5 gNa2O
do the same with O2
64.3 gO2 x (1 molO2/32.0gO2) x (2 molNa2O/1 mol O2) x (62.0gNa2O/1molNa2O) = 249.2 g Na2O
now you must pick the least amount of Na2O for the one that you actually get in the reaction. This is because you have to have both reacts still present for a reaction to occur. So after the Na runs out when it makes 75.5 gNa2O with O2, the reaction stops.
So, the mass of sodium oxide is
75.5 g</span>
To find the molar mass<span> of </span>Ba(NO3)2<span>, determine the </span>molar masses of all the atoms that form it. The Molar mass for Barium nitrate is <span>261.337 g/mol.</span>
Answer:
110L
Explanation:
Boyle's Law states that P1×V1=P2×V2
Volume is indirectly proportional to Pressure so P×V is constant
P1=55atm
V1=6L
P2=3atm
V2 is to be found
P1×V1=P2×V2
6×55=3×V2
330=3×V2
Answer: V2=110L
