Molarity = Moles of solute/ L(liters) of solution
So let's plug in the information.
5.0 moles/10L = 0.5 M
Answer:
balanced equation mole ratio 5 2 mol NO/1 mol O2
10.00 g O2 3 1 mol O2/32.00 g O2 5 0.3125 mol O2
20.00 g NO 3 1 mol NO/30.01 g NO 5 0.6664 mol NO
actual mole ratio 5 0.6664 mol NO/0.3125 mol O2 5 2.132 mol NO/1.000 mol O2
Because the actual mole ratio of NO:O2 is larger than the balanced equation mole
ratio of NO:O2, there is an excess of NO; O2 is the limiting reactant.
Mass of NO used 5 0.3125 mol O2 3 2 mol NO/1 mol O2 5 0.6250 mol NO
0.6250 mol NO 3 30.01 g NO/1 mol NO 5 18.76 g NO
Mass of NO2 produced 5 0.6250 mol NO2 3 46.01 g NO2/1 mol NO2 5 28.76 g NO2
Excess NO 5 20.00 g NO 2 18.76 g NO 5 1.24 g N
Explanation:
#1. An element or ion that has lost two electrons must have a net charge of 2+, because it has two more protons than electrons, therefore the answer is Mg2+
#2. aluminum ions have an oxidation state of 3+ and fluoride has an oxidation state of 1-, therefore I’m order for the charges to cancel you need 3 fluoride ions.
Therefore, the answer is AlF3
<span>2C2H6 + 7O2 = 4CO2 + 6H2O
</span>
According to the equation of the reaction of ethane combustion, ethane and carbon dioxide have following stoichiometric ratio:
n(C2H6) : n(CO2) = 1 : 2
n(CO2) = 2 x n(C2H6)
n(CO2) = 2 x 5.2 = 10.4 mole of CO2 is formed
