Respuesta :
The answer is: [B]: " 2 " .
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Explanation:
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Given the chemical equation:
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? H₂ + O₂ → 2 H₂O ;
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→ We are asked, "What coefficient, if any — should be put in front of the: " H₂ " ;
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(which is on the "left-hand side" side of the chemical equation given— the "reactants") ;
→ to get a balanced chemical equation?
__________________________________________________
→ Let us examine the "right-hand side" of this chemical equation—the product(s). In this case, the "product" given is: " 2 H₂O " .
So, on the "right hand side", we have:
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1) 4 "H's" → {Two "H₂" 's = 2 * 2 = "4 H's"} ; and:
2) 2 "O's" → { Two "O's").
__________________________________________________
So, the left-hand side should have:
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1) 4 "H's" ; and:
2) 2 "O's" ;
__________________________________________________
Let us examine the left-hand side (the "reactants").
__________________________________________________
" ? H₂ + O₂ " ;
__________________________________________________
On the left-hand side, we already have:
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1) " 2 O's " ; → one "O₂" = "2 O's" ; and:
2) "2 H's " ; → one "H₂" .
Now, we would need:
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A "total of "4 H's". Is there any number we could put as a coefficient on the other reactant, which happens to be: "H₂" ; to make a total of "4 H's" ?
{Note: There are only these TWO (2) reactants in this chemical equation.}.
→ The "H₂" ; as it stands alone, is insufficient—since that would be only "2 H's".
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→ Thus, we can rule out: "Answer choices: [A] and [D]."
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Note:
→ Choice [A]: "1" ; The coefficient, "1"; is generally not (never?) used; and basically would function as the same as:
→ Choice: [D]: "no coefficient is needed".
→ Choices [A] & [D]: would leave us with only "2 H's" on the "reactants side" (i.e. "left-hand side of the equation"; and we need FOUR ("4 H's").
__________________________________________________
Since we are given: "H₂" ; what coefficient could we put in front of this to get: "4 H's" ? (4÷2 =2). So we could put a "2" in front of the "H₂" ; to get:
"4 H's". The coefficient, "2" , corresponds directly with:
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→ Answer choice: [B]: "2" .
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{ NOTE: The remaining answer choice, [C], which is, "3" ; is INCORRECT; since 3 "H₂'s" would be "6 H's" ; (since 3 * "2 H's" equal "6 H's") ; which is too many "H's" → We need exactly "4 H's".}.
_________________________________________________
So, the correctly balanced equation is:
2 H₂ + O₂ → 2 H₂O ;
_________________________________________________
→ The coefficient that goes before the "H₂" ; to make this chemical equation balanced, is: "2" .
_________________________________________________
The correct answer is: [B]: "2" .
→ The coefficient that goes before the "H₂" is: "2" .
_________________________________________________
Hope this answer—and {lengthy} explanation—is of help!
_________________________________________________
_____________________________________________
Explanation:
____________________________________
Given the chemical equation:
____________________________________
? H₂ + O₂ → 2 H₂O ;
____________________________________________
→ We are asked, "What coefficient, if any — should be put in front of the: " H₂ " ;
__________________________________________________
(which is on the "left-hand side" side of the chemical equation given— the "reactants") ;
→ to get a balanced chemical equation?
__________________________________________________
→ Let us examine the "right-hand side" of this chemical equation—the product(s). In this case, the "product" given is: " 2 H₂O " .
So, on the "right hand side", we have:
__________________________________________________
1) 4 "H's" → {Two "H₂" 's = 2 * 2 = "4 H's"} ; and:
2) 2 "O's" → { Two "O's").
__________________________________________________
So, the left-hand side should have:
__________________________________________________
1) 4 "H's" ; and:
2) 2 "O's" ;
__________________________________________________
Let us examine the left-hand side (the "reactants").
__________________________________________________
" ? H₂ + O₂ " ;
__________________________________________________
On the left-hand side, we already have:
__________________________________________________
1) " 2 O's " ; → one "O₂" = "2 O's" ; and:
2) "2 H's " ; → one "H₂" .
Now, we would need:
__________________________________________________
A "total of "4 H's". Is there any number we could put as a coefficient on the other reactant, which happens to be: "H₂" ; to make a total of "4 H's" ?
{Note: There are only these TWO (2) reactants in this chemical equation.}.
→ The "H₂" ; as it stands alone, is insufficient—since that would be only "2 H's".
__________________________________________________
→ Thus, we can rule out: "Answer choices: [A] and [D]."
___________________________________________________
Note:
→ Choice [A]: "1" ; The coefficient, "1"; is generally not (never?) used; and basically would function as the same as:
→ Choice: [D]: "no coefficient is needed".
→ Choices [A] & [D]: would leave us with only "2 H's" on the "reactants side" (i.e. "left-hand side of the equation"; and we need FOUR ("4 H's").
__________________________________________________
Since we are given: "H₂" ; what coefficient could we put in front of this to get: "4 H's" ? (4÷2 =2). So we could put a "2" in front of the "H₂" ; to get:
"4 H's". The coefficient, "2" , corresponds directly with:
__________________________________________________
→ Answer choice: [B]: "2" .
_________________________________________________
{ NOTE: The remaining answer choice, [C], which is, "3" ; is INCORRECT; since 3 "H₂'s" would be "6 H's" ; (since 3 * "2 H's" equal "6 H's") ; which is too many "H's" → We need exactly "4 H's".}.
_________________________________________________
So, the correctly balanced equation is:
2 H₂ + O₂ → 2 H₂O ;
_________________________________________________
→ The coefficient that goes before the "H₂" ; to make this chemical equation balanced, is: "2" .
_________________________________________________
The correct answer is: [B]: "2" .
→ The coefficient that goes before the "H₂" is: "2" .
_________________________________________________
Hope this answer—and {lengthy} explanation—is of help!
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