Answer:
9.72 grams.
Explanation:
From the equation, 4 moles of NH₃ produce 6 moles of water.
Therefore the reaction to product ratio of NH₃ to H₂O is 4:6
and 2:3 into its simplest form.
The number of moles of NH₃ in 6.12 g is:
Number of moles=mass/ RMM
=6.12 g/17 G/mol
=0.36 moles.
Therefore the number of moles of H₂O produced is calculated as follows.
(0.36 Moles×3)2 = 0.54 moles
Mass= Number of moles × RMM
=0.54 moles×18g/mol
=9.72 grams.
Answer:
not great
Explanation:
my anxiety is acting up and im pretty sure i just flunked a bio test :/
how are you?
The excess reactant when 4.35 g of hydrogen reacts with 30.75g of hydrogen iodide is hydrogen
<h3>calculation</h3>
write the equation for the reaction
H2(g) + I2 (g) → 2HI (g)
find the mole of each reactant
moles = mass/molar mass
moles of H2= 4.35 g/2 g/mol= 2.175 moles
moles o I2 = 30.75 g/ 254 g/mo=0.1211 moles
0.1211 moles of I2 reacted with 0.1211 moles of H2 but there are more moles of H2 there Hydrogen was in excess
Answer:
![\large \boxed{\text{-486 kJ}}](https://tex.z-dn.net/?f=%5Clarge%20%5Cboxed%7B%5Ctext%7B-486%20kJ%7D%7D)
Explanation:
You calculate the energy required to break all the bonds in the reactants.
Then you subtract the energy needed to break all the bonds in the products.
2H₂ + O₂ ⟶ 2H-O-H
Bonds: 2H-H 1O=O 4H-O
D/kJ·mol⁻¹: 436 498 464
![\begin{array}{rcl}\Delta H & = & \sum{mD_{\text{reactants}}} - \sum{nD_{\text{products}}}\\& = & 2 \times 436 +1 \times 498 - 4 \times 464\\&=& 1370 - 1856\\&=&\textbf{-486 kJ}\\\end{array}\\\text{The enthalpy of reaction is $\large \boxed{\textbf{-486 kJ}}$}.](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7D%5CDelta%20H%20%26%20%3D%20%26%20%5Csum%7BmD_%7B%5Ctext%7Breactants%7D%7D%7D%20-%20%5Csum%7BnD_%7B%5Ctext%7Bproducts%7D%7D%7D%5C%5C%26%20%3D%20%26%202%20%5Ctimes%20436%20%2B1%20%5Ctimes%20498%20-%204%20%5Ctimes%20464%5C%5C%26%3D%26%201370%20-%201856%5C%5C%26%3D%26%5Ctextbf%7B-486%20kJ%7D%5C%5C%5Cend%7Barray%7D%5C%5C%5Ctext%7BThe%20enthalpy%20of%20reaction%20is%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B-486%20kJ%7D%7D%24%7D.)