Answer:
Moles of NO₂ = 0.158
Explanation:
SO 2 ( g ) + NO 2 ( g ) ⇄ SO 3 ( g ) + NO ( g )
According to the law of mass equation
= ![\frac{[SO_{3} ][NO]}{[SO_{2}][NO_{2} ]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BSO_%7B3%7D%20%5D%5BNO%5D%7D%7B%5BSO_%7B2%7D%5D%5BNO_%7B2%7D%20%20%5D%7D)
⇒ 3.10 = ![\frac{(1.00)(1.00)}{(2.30) [NO_{2} ]}](https://tex.z-dn.net/?f=%5Cfrac%7B%281.00%29%281.00%29%7D%7B%282.30%29%20%5BNO_%7B2%7D%20%5D%7D)
At equilibrium [SO₃] = [NO]
⇒ [NO₂] = 
⇒ [NO₂] = 0.158
So. number of moles of NO₂ at equilibrium added = 0.158
O1Fl2
1. Assume an 100g sample, so the percentage will stay the same
2. Covert each element into their molar mass
29.6/16.00=1.8 mols of O
70.4/19.00=3.7 mols of Fl
3. Divide both by the smallest value of mol
1.8/1.8=1 O
3.7/1.8=2 Fl
4. Write the empirical formula:
O1Fl2
According to the law of conservation of Mass:
In a chemical reaction mass can neither be created nor be destroyed
So, we can say that: Mass of A + Mass of B = Mass of C
In the given reaction,
One of the reactants weigh 5 grams and another one weighs x grams
The mass of the product of this reaction is 9 grams
<u>Mass of reactant B:</u>
Mass of A + Mass of B = Mass of C
5 + x = 9
x = 4 grams
Answer:
12.44 g
Explanation:
2C4H10 + 13O2 = 8CO2 + 10H2O
n(C4H10) = m(C4H10)/M(C4H10) = 4.1 / 58g/mol = 0.0707 mol (excess).
n(O2) = m(O2)/M(O2) = 25.9 / 32g/mol = 0.809 mol (deficiency).
Since the ratio of O2 to octane is 13 : 2 we can divide 0.0707 by 2 to get 0.03535 and divide 0.809 by 13 to get 0.062.
mass of CO2 produced =
M = [0.0707 moles C4H10 x 8 moles CO2] / 2 moles C4H10 x 44 g CO2/mol
M = 0.5656/2 * 44
M = 0.2828 * 44
M = 12.44 of CO2
