<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>
Answer:
Its mass is measured on a scale.
It is measured on a balance.
Explanation:
The mass of an object can easily be determined using a balance or weighing scale.
Mass is the amount of matter contained in a substance. it is has the same value every where and will not vary by geography or location.
Weight is the force on body due to gravity. It is a function of mass and acceleration due to gravity.
Both mass and weight are different.
They can both be measured using a weight device.
Firstly calculate the grams in the last 8 percent before moving onto the pyrite section.
50.8x0.08=4.064g
We know that iron ore in this case has 92 percent pyrite which contains 46.5 percent iron so we do 50.8x0.92=46.736g from this we need to find 46.5 percent of the iron content in the 92 percent pyrite section then add this answer to the 8 percent of iron ore we found at the start.46.736x0.465=21.73224g
21.73224g+4.064=25.79624g of iron ore 25.8g(3sf)
Answer:
A. There was still 140 ml of volume available for the reaction
Explanation:
According to Avogadro's law, we have that equal volumes of all gases contains equal number of molecules
According to the ideal gas law, we have;
The pressure exerted by a gas, P = n·R·T/V
Where;
n = The number of moles
T = The temperature of the gas
R = The universal gas constant
V = The volume of the gas
Therefore, given that the volumes and number of moles of the removed air and added HCl are the same, the pressure and therefore, the volume available for the reaction will remain the same
There will still be the same volume available for the reaction.
Answer:
0.719 moles of NH₃
Explanation:
Molar mass of ammonia 17 g/mol
Mass of amonia = 12.23 g
Mass / Molar mass = Moles
12.23 g / 17 g/mol = 0.719 moles
