Answer:
16.9g of H₂O can be formed
Explanation:
Based on the chemical reaction, 2 moles of H₂ react per mole of O₂. To anser this question we must find limiting reactant converting the mass and volume of each reactant to moles:
<em>Moles H₂ -Molar mass: 2.016g/mol-:</em>
8.76g * (1mol / 2.016g) = 4.345 moles
<em>Moles O₂:</em>
PV = nRT
PV/RT = n
P = 1atm at STP
V = 10.5L
R = 0.082atmL/molK
T = 273.15K at STP
n = 1atm*10.5L / 0.082atmL/molK*273.15K
n = 0.469 moles of oxygen
For a complete reaction of 4.345 moles moles of hydrogen are required:
4.345 moles H2 * (1mol O2 / 2mol H2) = 2.173 moles of O2 are required. As there are just 0.469 moles, Oxygen is limiting reactant
Now, 1 mole of O2 produce 2 moles of H2O. 0.469 moles will produce:
0.469 moles O₂ * (2 moles H₂O / 1mol O₂) = 0.938 moles H₂O.
The mass is -Molar mas H₂O = 18.01g/mol-:
0.938 moles * (18.01g/mol) =
<h3>16.9g of H₂O can be formed</h3>
Answer:
N2O2(g) +O2(g) ===> 2NO2(g)
Explanation:
For a nonelementary reaction, the reaction equation is described as the sum of all the steps involved. All these steps constitute the reaction mechanism. Each step in the mechanism is an elementary reaction. The rate law of the overall reaction involves the rate determining step (slowest step) in the reaction sequence.
Now look at the overall reaction 2NO(g) + O2(g) ---------> 2NO2(g)
The two steps in the mechanism are
2NO(g) --------->N2O2(g) (fast)
N2O2(g) +O2(g) -------> 2NO2(g) (slow)
Summing all the steps and cancelling out the intermediate N2O2(g), we obtain the reaction equation;
2NO(g) + O2(g) ---------> 2NO2(g)
Hence the answer.
I hope this picture helps
