<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:
Explanation:
You have the equation. Now change the 3.4 g H2 to moles. moles = grams/molar mass
<em>3.4 g/2.016 = 1.686 moles.</em>
Now using the coefficients in the balanced equation, convert moles H2O2 to moles H2O.
1.686 moles H2 x (2 moles H2O/2 moles H2O2) = 1.686 x (2/2) = 1.686 x (1/1) = 1.686 moles H2O.
Now you know that 1 mole of water is composed of 6.022 x 10^23 molecules. So
1.686 moles H2O x (6.022 x 10^23 molecules H2O/1 mole H2O) = ?? molecules.
Elements in the same group share the same number of valence electrons.
The precipitation reaction of sodium nitrate and lead (II) chloride would not happen. No reaction will happen between these substances. Hope this answers the question. Have a nice day.
