A 1.50 L portion of 0.100M sulfamic acid, HSO3NH2, is placed in the bottom of a 10.0 L glass lined stainless steel chamber. A dish containing 17.25g of solid NaNO2 is floated on the acid. A thin open ended U tube containing a pool of liquid mercury is connected to the chamber with a short length of gas-tight tubing. When the assembly is prepared as shown, the chamber is gently shaken to overturn the floating dish. The resulting reaction proceeds to completion and returns to room temperature.
HSO3NH2 (aq) + NaNO2 (s) = H2O (l) + N2 (g) + NaHSO4 (aq)
1) What mass of N2 will be produced by the complete consumption of the sulfamic acid?
a) 7.00 g
b) 4.20 g
c) 2.80 g
d) 0.00210 g
To determine the mass of N2 produced by the complete consumption of the sulfamic acid, we need to use stoichiometry.
First, let's find the number of moles of sulfamic acid (HSO3NH2) in the given 1.50 L of 0.100 M solution:
moles of HSO3NH2 = volume of solution (L) x concentration of solution (M)
moles of HSO3NH2 = 1.50 L x 0.100 M = 0.150 moles
From the balanced chemical equation, we can see that the stoichiometric ratio between HSO3NH2 and N2 is 1:1. This means that for every 1 mole of HSO3NH2 consumed, 1 mole of N2 is produced.
Therefore, the number of moles of N2 produced will also be 0.150 moles.
Now, we need to find the molar mass of N2:
Molar mass of N2 = 2 x atomic mass of Nitrogen
Molar mass of N2 = 2 x 14.01 g/mol = 28.02 g/mol
Finally, we can calculate the mass of N2 produced:
mass of N2 = moles of N2 x molar mass of N2
mass of N2 = 0.150 moles x 28.02 g/mol = 4.203 g
Therefore, the mass of N2 produced by the complete consumption of sulfamic acid is approximately 4.20 g.
So, the correct answer is b) 4.20 g.