The limiting reactant when 5.6 moles of aluminium react with 6.2 moles of water is
water( H2O)
<u><em>Explanation</em></u>
The balanced equation is as below
2 Al +3 H2O → Al2O3 +3 H2
The mole ratio of Al :Al2O3 is 2:1 therefore the moles of Al2O3
= 5.6 x1/2 = 2.8 moles
The mole ratio of H2O: Al2O3 is 3:1 therefore the moles of Al2O3 produced
= 6.2 x1/3= 2.067 moles
since H2O yield less amount of Al2O3 , H2O is the limiting reagent.
BF3 .... BP = −100.3 °C
<span>RbCl ..... solid </span>
<span>CH3SCH3 ..... BP = 35-41 °C </span>
<span>SbH3 .... BP = −17 °C </span>
<span>SiS2 ..... solid </span>
<span>Ethanol solid --> ethanol melts --> ethanol liquid </span>
<span>-135C ---------------> -114C --------------> -50C </span>
<span>............ ΔT = 21C ....... ....... ΔT = 64C </span>
Answer:
D
Explanation:
A single-replacement reaction occurs when one element replace another in a single compound.
ΔH for this reaction : -2668.4 (exothermic)
<h3>Further explanation </h3>
The change in enthalpy in the formation of 1 mole of the elements is called enthalpy of formation
The enthalpy of formation measured in standard conditions (25 ° C, 1 atm) is called the standard enthalpy of formation (ΔHf °)
(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)
The value of ° H ° can be calculated from the change in enthalpy of standard formation:
∆H ° rxn = ∑n ∆Hf ° (product) - ∑n ∆Hf ° (reactants)
you can search the value of ΔHf on the internet
∆H ° rxn =( ∆H Al₂O₃+∆H AlCl₃+3.∆H NO + 6.∆H H₂O)-(∆H Al+∆H NH₄ClO₄)
The problem in this question can be solved using the ideal gas formula. Ideal gas formula show interaction between the pressure, volume, temperature and the number of molecules.
n= number of molecule/ avogadro number
n= (2.0116 x 10^23 / 6.02*10^23)= 1/3 mole
T= celcius + 273.15
T= (41+273.15) K=314.15
P=2280 mmHg / (760mmHg/atm)= 3 atm
PV=nRT
V= nrT/ P
V= 1/3 moles * (<span>0.08205 L atm / mol·K) </span> * 314.15 K / 3 atm
V= 2.864 L