How many grams of nitric acid(HNO3) will be produced when 0.51 mol of dinitrogen pentoxide reacts with water to form nitric acid?
mol mass N2O5 = 2*14 + 5*16 = 108 g/mol
mol mass HNO3 = 1 + 14 + 3*16 = 63 g/mol
for every N2O5 molecule you get two HNO3 molecules to balance the N atoms
so
.51 mol N2O5 ---> 1.02 mol HNO3
1.02mol * 63 g/mol = 64.26 g
To find the number of grams of nitric acid (HNO3) produced when 0.51 mol of dinitrogen pentoxide reacts with water, you'll need to use the molar ratio between dinitrogen pentoxide (N2O5) and nitric acid (HNO3).
The balanced chemical equation for the reaction is:
N2O5 + H2O -> 2HNO3
From the equation, you can see that for every 1 mol of N2O5, 2 mol of HNO3 are produced. Therefore, you can write the following equation to find the number of moles of HNO3 produced:
moles of HNO3 = moles of N2O5 * (2 mol HNO3 / 1 mol N2O5)
Now, substitute the given value:
moles of HNO3 = 0.51 mol * (2 mol HNO3 / 1 mol N2O5)
moles of HNO3 = 1.02 mol HNO3
Finally, use the molar mass of HNO3 to convert moles to grams:
molar mass of HNO3 = 1 * molar mass of H + 1 * molar mass of N + 3 * molar mass of O
= 1 * 1.01 g/mol + 1 * 14.01 g/mol + 3 * 16.00 g/mol
= 63.01 g/mol
grams of HNO3 = moles of HNO3 * molar mass of HNO3
= 1.02 mol * 63.01 g/mol
= 64.2662 g (rounded to four decimal places)
Therefore, approximately 64.27 grams of nitric acid (HNO3) will be produced.
To find the number of grams of nitric acid (HNO3) produced, we need to use the molar ratio between dinitrogen pentoxide (N2O5) and nitric acid.
First, let's write the balanced chemical equation for the reaction between dinitrogen pentoxide and water to form nitric acid:
N2O5 + H2O -> 2HNO3
From the balanced equation, we can see that 1 mole of N2O5 produces 2 moles of HNO3. This means that the molar ratio between N2O5 and HNO3 is 1:2.
Now, we can calculate the number of moles of nitric acid produced using the given number of moles of dinitrogen pentoxide:
0.51 mol N2O5 * (2 mol HNO3 / 1 mol N2O5) = 1.02 mol HNO3
Next, we need to calculate the mass of nitric acid using its molar mass. The molar mass of HNO3 is:
H = 1.01 g/mol
N = 14.01 g/mol
O = 16.00 g/mol
Molar mass of HNO3 = (1.01 g/mol) + (14.01 g/mol) + (3 * 16.00 g/mol) = 63.01 g/mol
Finally, we can calculate the mass of nitric acid produced:
Mass of HNO3 = number of moles * molar mass = 1.02 mol * 63.01 g/mol = 64.27 g
Therefore, approximately 64.27 grams of nitric acid (HNO3) will be produced when 0.51 moles of dinitrogen pentoxide reacts with water to form nitric acid.