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.
Plants, animals and bacteria
The root words in this item are hydro- are lysis-. Hydro pertains to water and lysis is the disintegration of a cell or a compound. The word hydrolysis is the disintegration or the breaking of the compound because or due to its reaction with water. This usually brings about changes in pH of a solution.
2 C₃H₇OH (l) + 9 O₂ (g) → 6 CO₂ (g) + 8 H₂O (g)
Explanation:
To balance the chemical equation the number of atoms of each element entering the reaction have to be equal to the number of atoms of each element leaving the reaction, in order to conserve the mass.
Bellow we have the balanced chemical equation of the complete combustion of C₃H₇OH:
C₃H₇OH (l) + (9/2) O₂ (g) → 3 CO₂ (g) + 4 H₂O (g)
to have integer coefficients we multiply the reaction with 2:
2 C₃H₇OH (l) + 9 O₂ (g) → 6 CO₂ (g) + 8 H₂O (g)
where:
(l) - liquid
(g) - gaseous
Learn more about:
combustion reaction
brainly.com/question/9425444
balancing chemical equations
brainly.com/question/13941483
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Answer:
63.05% of MgCO3.3H2O by mass
Explanation:
<em>of MgCO3.3H2O in the mixture?</em>
The difference in masses after heating the mixture = Mass of water. With the mass of water we can find its moles and the moles and mass of MgCO3.3H2O to find the mass percent as follows:
<em>Mass water:</em>
3.883g - 2.927g = 0.956g water
<em>Moles water -18.01g/mol-</em>
0.956g water * (1mol/18.01g) = 0.05308 moles H2O.
<em>Moles MgCO3.3H2O:</em>
0.05308 moles H2O * (1mol MgCO3.3H2O / 3mol H2O) =
0.01769 moles MgCO3.3H2O
<em>Mass MgCO3.3H2O -Molar mass: 138.3597g/mol-</em>
0.01769 moles MgCO3.3H2O * (138.3597g/mol) = 2.448g MgCO3.3H2O
<em>Mass percent:</em>
2.448g MgCO3.3H2O / 3.883g Mixture * 100 =
<h3>63.05% of MgCO3.3H2O by mass</h3>
