Answer:
12 moles H
2
O
Explanation:
Your tools of choice for stoichiometry problems will always be the mole ratios that exist between the chemical species that take part in the reaction.
As you know, the stoichiometric coefficients attributed to each compound in the balanced chemical equation can be thought of as moles of reactants needed or moles of products formed in the reaction.
In your case, the balanced chemical equation for this synthesis reaction looks like this
2
H
2(g]
+
O
2(g]
→
2
H
2
O
(l]]
Notice that the reaction requires
2
moles of hydrogen gas and
1
mole of oxygen gas to produce
2
moles of water.
This tells you that the reaction produces twice as many moles of water as you have moles of oxygen gas that take part in the reaction.
You know that your reaction uses
6.0
moles of oxygen. Assuming that hydrogen gas is not a limiting reagent, you can say that the reaction will produce
6.0
moles O
2
⋅
2
moles H
2
O
1
moles O
2
=
12 moles H
2
O
Explanation:
- C2H4 + H20 --> C2H5O
- <span>C3H8 + Cl2--> C3H7Cl +HCl
- </span><span>C2H2+Br2 --> C2H2Br2</span>
- C4H10+Br2 --> C4H10+HBr
- <span>C3H6 + BR2 --</span>> C3H6BR2
The answer is going to be b. hope that helped
Substance combines with another to form a new substance is a chemical change
The question is incomplete; the complete question is;
What alkene would give the products below after reaction with O3, followed by reduction with (CH3)2S? Write the condensed structural formula (CH3)2C=O and CH3-CH2-CH=O
Answer:
2-methylpent-2-ene (C6H12)
Explanation:
Ozonolysis is a wonderful method for determining the location of double bonds in an alkene since the oxygenated carbons in the carbonyl compounds formed after ozonolysis are the ones that were initially joined by the double bonds in the original alkene.
Hence if an alkene yields (CH3)2C=O and CH3-CH2-CH=O, the original alkene must be C6H12, that is, 2-methylpent-2-ene