Answer:
The answer to your question is
Explanation:
Data
12.5 g of reactant
Balanced Reaction 1
TiBr₄ + 2H₂ ⇒ Ti + 4HBr
Molar mass of TiBr₄ = 48 + (4 x 80) = 368 g
Atomic mass of Ti = 48 g
Molar mass of HBr = 1 + 80 = 81
368 g of TiBr₄ ---------------- 48 g of Ti
12.5 g of TiBr₄ -------------- x
x = (12.5 x 48) / 368
x = 1.63 g of Ti
368 g of TiBr₄ ----------------4(81) g of HBr
12.5 g of TiBr₄ ------------- x
x = (12.5 x 324) / 368
x = 11 g of HBr
Balanced reaction 2
3SiH₄ + 4NH₃ ⇒ Si₃N₄ + 12H₂
Molar mass of SiH₄ = 28 + 4 = 32
Molar mass of Si₃N₄ = 28 x 3 + 14 x 4 = 84 + 56 = 140 g
Atomic mass of H₂ = 2 g
3(32) g of SiH₄ --------------- 140 g of Si₃N₄
12.5 g of SiH₄ -------------- x
x = 18.2 g of Si₃N₄
3(32) g of SiH₄ --------------- 24 g of H₂
12.5 g of SiH₄ -------------- x
x = 3.125 g of H₂
It is quite easy:
1 cal = 4,1868 J
Solution is:
2930 kcal = 4,1868 * 2930 [kJ] = 12267,324 [kJ]
Answer:
Explanation:
2Al(s) + Fe₂O₃(s) ⟶ Al₂O₃(s) + 2Fe(s); ΔᵣH = ?
The formula for calculating the enthalpy change of a reaction by using the enthalpies of formation of reactants and products is
2Al(s) + Fe₂O₃(s) ⟶ Al₂O₃(s) + 2Fe(s)
ΔfH°/kJ·mol⁻¹: 0 -824.3 -1675.7 0
![\begin{array}{rcl}\Delta_{\text{r}}H^{\circ} & = & [1(-1675.7) + 2(0)] - [2(0) - 1(-824.3)]\\& = & -1675.7 + 824.3\\& = & \textbf{-851.4 kJ/mol}\\\end{array}\\\text{The enthalpy change is } \large \boxed{\textbf{-851.4 kJ/mol}}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7D%5CDelta_%7B%5Ctext%7Br%7D%7DH%5E%7B%5Ccirc%7D%20%26%20%3D%20%26%20%5B1%28-1675.7%29%20%2B%202%280%29%5D%20-%20%5B2%280%29%20-%201%28-824.3%29%5D%5C%5C%26%20%3D%20%26%20-1675.7%20%2B%20824.3%5C%5C%26%20%3D%20%26%20%5Ctextbf%7B-851.4%20kJ%2Fmol%7D%5C%5C%5Cend%7Barray%7D%5C%5C%5Ctext%7BThe%20enthalpy%20change%20is%20%7D%20%5Clarge%20%5Cboxed%7B%5Ctextbf%7B-851.4%20kJ%2Fmol%7D%7D)
