Respuesta :
Answer:
The answer to your question is Only A and B
Explanation:
Reactions
A. 3 AgBr + GaPO₄ → Ag₃PO₄ + GaBr₃
Reactants Elements Products
3 Ag 3
3 Br 3
1 Ga 1
1 P 1
4 O 4
B. 3H₂SO₄ + 2B(OH)₃ → B₂(SO₄)₃ + 6H₂O
Reactants Elements Products
3 S 3
2 B 2
12 H 12
18 O 18
C. Fe + 2 AgNO₃ → Fe(NO₃)₂ + 2 Ag
Reactants Elements Products
1 Fe 1
2 Ag 2
2 N 2
6 O 6
D. C₂H₄O₂ + 2O₂ → 2CO₂ + 2H₂O
Reactants Elements Products
2 C 2
4 H 4
6 O 6
Answer:
In option A and B we have a reactant with coefficient 3 in front
Explanation:
A. AgBr + GaPO4 → Ag3PO4 +GaBr3
On the left side we have 1x Br (in AgBr), on the right side we have 3x Br (in GaBr3). To balance the amount of Br on both sides, we have to multiply AgBr (on the left side) by 3. Now we also have balanced the amount of Ag on both sides. The equation is balanced now.
3AgBr + GaPO4 → Ag3PO4 +GaBr3
B. H2SO4 + B(OH)3 → B2(SO4)3 + H2O
On the left side we have 1x S (in H2SO4) on the right side we have 3x S (inB2(SO4)3. To balance the amount of S on both sides, we have to multiply H2SO4, on the left side, by 3.
3H2SO4 + B(OH)3 → B2(SO4)3 + H2O
On the left side we have 6x H (in 3H2SO4) on the right side we have 1x H (in H2O). To balance the amount of H on both sides, we have to multiply H2O, on the right side, by6.
3H2SO4 + B(OH)3 → B2(SO4)3 + 6H2O
On the left side we have 1x B (in B(OH)3) on the right side we have 2x B (in B2(SO4)3). To balance the amount of B on both sides, we have to multiply B(OH)3, on the leftt side, by 2.
Now the equation is balanced.
3H2SO4 + 2B(OH)3 → B2(SO4)3 + 6H2O
C. Fe + AgNO3 → Fe(NO3)2 + Ag
On the left side we have 1x NO3 (in AgNO3), on the right side we have 2x NO3 (in Fe(NO3)2). To balance the amount of NO3 on both sides, we have to multiply AgNO3 (on the left side) by 2
Fe + 2AgNO3 → Fe(NO3)2 + Ag
On the left side we have 2x Ag (in 2AgNO3), on the right side we have 1x Ag . To balance the amount of Ag on both sides, we have to multiply Ag (on the right side) by 2
Fe + 2AgNO3 → Fe(NO3)2 + 2Ag
D. C2H4O2 + O2 → CO2 +H2O
On the left side we have 2x C (in C2H4O2), on the right side we have 1x C (in CO2) . To balance the amount of C on both sides, we have to multiply CO2 (on the right side) by 2
C2H4O2 + O2 → 2CO2 +H2O
On the left side we have 4x H (in C2H4O2), on the right side we have 2x H (in H2O) . To balance the amount of H on both sides, we have to multiply H2O (on the right side) by 2
C2H4O2 + O2 → 2CO2 + 2H2O
