Answer:
The answer to your question is V = 0.32 L
Explanation:
Data
Volume of NH₃ = ?
P = 3.2 atm
T = 23°C
mass of CaH₂ = 2.65 g
Balanced chemical reaction
6Ca + 2NH₃ ⇒ 3CaH₂ + Ca₃N₂
Process
1.- Convert the mass of CaH₂ to moles
-Calculate the molar mass of CaH₂
CaH₂ = 40 + 2 = 42 g
42 g ------------------ 1 mol
2.65 g -------------- x
x = (2.65 x 1)/42
x = 0.063 moles
2.- Calculate the moles of NH₃
2 moles of NH₃ --------------- 3 moles of CaH₂
x --------------- 0.063 moles
x = (0.063 x 2) / 3
x = 0.042 moles of NH₃
3.- Convert the °C to °K
Temperature = 23°C + 273
= 296°K
4.- Calculate the volume of NH₃
-Use the ideal gas law
PV = nRT
-Solve for V
V = nRT / P
-Substitution
V = (0.042)(0.082)(296) / 3.2
-Simplification
V = 1.019 / 3.2
-Result
V = 0.32 L
At the end of the reaction, the catalyst is UNCHANGED.
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Answer:
The amount in grams of hydrogen gas produced is 0.551 grams
Explanation:
The parameters given are;
Number of atoms of potassium, aₙ = 3.289 × 10²³ atoms
Chemical equation for the reaction is given as follows;
2K + 2H₂O 
KOH + H₂
Avogadro's number, 
, regarding the number of molecules or atom per mole is given s follows;
= 6.02 × 10²³ atoms/mole
Therefore;
The number of moles of potassium present = 3.289 × 10²³/(6.02 × 10²³) = 0.546 moles
2 moles of potassium produces one mole of hydrogen gas, therefore;
1 moles of potassium produces 1/2 mole of hydrogen gas, and 0.546 moles of potassium will produce 0.546/2 moles of hydrogen which is 0.273 moles of hydrogen gas
The molar mass of hydrogen gas = 2.016 grams
Therefore, 0.273 moles will have a mass of 0.273×2.016 = 0.551 grams.
The amount in grams of hydrogen gas produced = 0.551 grams.
1. 2 H2 + O2 = 2 H2O
2. 6 K + B2O3 = 3 K2O + 2 B
3. 10 Na + 2 NaNO3 = 6 Na2O + N2
