The reaction is:
NH4 (NO3) (s) ⇄ N2O (g) + 2 H2O (g)
This means that 1 mol of NH4 (NO3)s produces 3 moles of gases.
Now find the number of moles in 1.71 kg of NH4 (NO3)
Molar mass = 2*14g/mol + 4 * 1g/mol + 3*16g/mol = 80 g/mol
# moles = mass / molar mass = 1710 g / 80 g/mol = 21.375 mol of NH4(NO3)
We already said that every mol of NH4(NO3) produces 3 moles of gases, then the number of moles of gases produced is 3 * 21.375 = 64.125 mol
Now use the equation for ideal gases to fin the volume
pV = nRT => V = nRT / p = (64.125 mol)(0.082atm*liter / K*mol) * (119 +273)K / (731mmHg *1 atm/760mmHg) =
V = 2143.01 liters
The answer has to be 22.4L
Answer:
55.0g water can be made
Explanation:
To solve this question, we must convert the molecules of H2 to moles using Avogadro's constant. With the moles, and the reaction:
H2 + 1/2O2 → H2O
We can find the moles of H2O = Moles H2 and its mass of using molar mass of water -H2O = 18.01g/mol-
<em>Moles H2 = Moles H2O:</em>
1.84x10²⁴ molecules * (1mol / 6.022x10²³ molecules) = 3.055 moles H2O
<em>Mass:</em>
3.055 moles H2O * (18.01g / mol) = 55.0g water can be made
