Na + NaNO3 = Na2O + N2
4 Na + 2 NaNO3 = 6 Na2O + N2
6 Na on each side
2 N on each side
6 O on each side
Answer:- 448 mL of hydrogen gas are formed.
Solution:- It asks to calculate the volume of hydrogen gas formed in milliliters at STP when 0.020 moles of magnesium reacts with excess HCl acid. The balanced equation is:

There is 1:1 mol ratio between Mg and hydrogen gas. So, the moles of hydrogen gas is also equals to the moles of Mg reacted.
moles of Hydrogen gas formed = 0.020 mol
At STP, volume of 1 mol of the gas is 22.4 L. We need to calculate the volume of 0.02 moles of hydrogen gas.

= 0.448 L
They want answer in mL. So, let's convert L to mL using the conversion formula, 1L = 1000mL

= 448 mL
So, 0.020 moles of magnesium would produce 448 mL of hydrogen gas at STP on reacting with excess of HCl acid.
Answer:
212.5 mL
both the original and the diluted solution have 0.765 moles of KCl
Explanation:
c1V1 = c2V2
V2 = c1V1/c2 = (1.8 M×425 mL)/1.2 M = 637.5 mL
(637.5 - 425) mL = 212.5 mL
n = (1.8 mol/L)(0.425 L) = 0.765 moles of KCl
since it's a dilution, the diluted solution has the same number of moles as the original solution, 0.765 moles of KCl