Burning wood, rusting of iron, and baking a cake
Answer:
46 g
Explanation:
The balanced equation of the reaction between O and NO is
2 NO + O₂ ⇔ 2 NO₂
Now, you need to find the limiting reagent. Find the moles of each reactant and divide the moles by the coefficient in the equation.
NO: (80 g)/(30.006 g/mol) = 2.666 mol
(2.666 mol)/2 = 1.333
O₂: (16 g)/(31.998 g/mol) = 0.500 mol
(0.500 mol)/1 = 0.500 mol
Since O₂ is smaller, this is the limiting reagent.
The amount of NO₂ produced will depend on the limiting reagent. You need to look at the equation to determine the ratio. For every mole of O₂ reacted, 2 moles of NO₂ are produced.
To find grams of NO₂ produced, multiply moles of O₂ by the ratio of NO₂ to O₂. Then, convert moles of NO₂ to find grams.
0.500 mol O₂ × (2 mol NO₂/1 mol O₂) = 1.000 mol NO₂
1.000 mol × 46.005 g/mol = 46.005 g
You will produce 46 g of NO₂.
Option B is correct
K = Kp /Kr
The given equation indicating, the product containing 6 moles of proton whereas the reactant contains 2 mole of bismuth and 3 mole of hydrogen sulphide.
Hence, in reaction B there are 2 mole of bismuth and 3 mole of hydrogen sulphide reacting to produce 6 moles of proton. whereas the concentration of Bi2S3 is not considered as it is present in solid phase.
The conversion of volume to moles at STP is 1 mole.
The ideal gas equation is given as :
P V = n R T
where,
P = pressure of the gas
V = volume of the gas
n = ?
R = constant = 0.823 atm L / mol K
T = temperature
At STP , the pressure is 1 atm and the temperature is 273.15 K, the volume At STP is 22.4 L.
moles , n = P V / R T
n = ( 1 × 22.4 ) / (0.0823 × 273.15)
n = 1 mole
Thus, at STP , the number of moles is 1 mol.
To learn more about moles here
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