Solving stoichiometry problems that involve mass Now that you know the molar masses of the relevant compounds, you are ready to start solving stoichiometry problems. In general, the typical strategy is Convert from grams of compound X to moles of compound X using the molar mass of compound X. Convert from moles of compound X to moles of compound Y using the coefficients in the balanced chemical equation. Convert from moles of compound Y to grams of compound Y using the molar mass of compound Y. Part B Calculate the mass of water produced when 9.56 g of butane reacts with excess oxygen.
Answers
14.85 g of water
Explanation:
Calculate the mass of water produced when 9.56 g of butane reacts with excess oxygen.
Here we have the full oxidation of butane (C₄H₁₀) which produce carbon dioxide and water.
C₄H₁₀ + (13/2) O₂ → 4 CO₂ + 5 H₂O
number of moles = mass / molecular weight
number of moles of butane = 9.56 / 58 = 0.165 moles
Now, taking in account the chemical reaction, we devise the following reasoning:
if 1 mole of butane produces 5 moles of water
then 0.165 moles of butane produces X moles of water
X = (0.165 × 5) / 1 = 0.825 moles of water
mass = number of moles × molecular weight
mass of water = 0.825 × 18 = 14.85 g
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oxidation of hydrocarbons
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Answer:
14.85 g of water
Explanation:
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Answers
Please help?????
Answers
Explanation:
4.92*10-3=4.92/10^3=
4.92/1000=0.00492
Answers
Answer:
1.4 g H₂O
Explanation:
In a reaction, the reactants are usually not present in exactstoichiometric amounts, that is, in the proportions indicated by the balanced equation. Frequently a large excess of one reactant is supplied to ensure that the more expensive reactant is completely converted to the desired product. Consequently, some reactant will be left over at the end of the reaction. The reactant used up first in a reaction is called the limiting reagent, because the maximum amount of product formed depends on how much of this reactant was originally present. When this reactant is used up, no more product can be formed.
CH3OH HNO2
N2 CH2O
Answers
Answer : The Lewis-dot structure of the following molecules are shown below.
Explanation :
Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.
In the Lewis-dot structure the valance electrons are shown by 'dot'.
(1) The given molecule is,
As we know that hydrogen has '1' valence electron, nitrogen has '5' valence electrons.
Therefore, the total number of valence electrons in = 5 + 3(1) = 8
According to Lewis-dot structure, there are 6 number of bonding electrons and 2 number of non-bonding electrons.
(2) The given molecule is,
As we know that sulfur and oxygen has '6' valence electrons.
Therefore, the total number of valence electrons in = 6 + 2(6) = 18
According to Lewis-dot structure, there are 8 number of bonding electrons and 10 number of non-bonding electrons.
(3) The given molecule is,
As we know that carbon has '4' valence electrons, hydrogen has '1' valence electron and oxygen has '6' valence electrons.
Therefore, the total number of valence electrons in = 4 + 4(1) + 6 = 14
According to Lewis-dot structure, there are 10 number of bonding electrons and 4 number of non-bonding electrons.
(4) The given molecule is,
As we know that hydrogen has '1' valence electron, nitrogen has '5' valence electrons and oxygen has '6' valence electrons.
Therefore, the total number of valence electrons in = 1 + 5 + 2(6) = 18
According to Lewis-dot structure, there are 8 number of bonding electrons and 10 number of non-bonding electrons.
(5) The given molecule is,
As we know that nitrogen has '5' valence electrons.
Therefore, the total number of valence electrons in = 2(5) = 10
According to Lewis-dot structure, there are 6 number of bonding electrons and 4 number of non-bonding electrons.
(6) The given molecule is,
As we know that carbon has '4' valence electrons, hydrogen has '1' valence electron and oxygen has '6' valence electrons.
Therefore, the total number of valence electrons in = 4 + 2(1) + 6 = 12
According to Lewis-dot structure, there are 8 number of bonding electrons and 4 number of non-bonding electrons.
2Cl-(aq)--->Cl2(g)+2e-
Cr3+(aq)+3e- ---->Cr(s)
What is the final, balanced equation for this reaction?
1.) 2cr3+(aq)+6Cl-(aq) ------> 2Cr(s)+3Cl2(g)
2.) 2Cr3(aq)+2Cl-(aq)+6e- --->Cl2(g)+2Cr(s)
3.) Cr3+(aq)+6Cl-(aq)+3e- ---->2Cr(g)+3Cl2(g)
4.) Cr3+(aq)+2Cl-(aq)------>Cr(s)+Cl2(g)
Answers
Explanation:
1) Write the oxidation half-reaction:
2) Write the reduction half-raction:
3) Multiply each half-reaction by the appropiate coefficient to equal the number of electrons of both half-reactions.
4) Add both half-reactions
And that is the answer. You can count the atoms and charges on every side and check they are equal.
Answer:
A
Explanation:
got it correct on edge
Answers
Answer:
H2O (water)
N2 (nitrogen)
O3 (ozone)
CaO (calcium oxide)
Explanation:
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