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Answers:3 KOH + Cr(NO3)3 = 3 KNO3 + Cr(OH)3 Zn + Cu(NO3)2 = Zn(NO3)2 + Cu 3 AgNO3 + K3PO4 = 3 KNO3 + Ag3PO4 Mg + H2SO4 = MgSO4 + H2 2Na + 2H2O =2 NaOH + H2 Pb(NO3)2 + 2 HCl = PbCl2 + 2 HNO3 2 KClO4 + H2SO4 = K2SO4 + 2 HClO4 Ba(OH)2 + 2 HBr = BaBr2 + 2 H2O 2 H3PO4 + 3 MgCl2 = Mg3(PO4)2 + 6 HCl NH4OH + NaClO = NaOH + NH4ClO CH3COOH + NH4OH = CH3COONH4 + H2O Ba(ClO3)2 + Li2CO3 = BaCO3 + 2 LiClO3
Answers:Balancing chemical equations is a difficult skill to master. I will try to explain it as well as I can: Let's start with aerobic respiration: C6H12O6 + O2 --> H2O +CO2 I am going to name the coefficients to make it easier to follow. (A) C6H12O6 + (B) O2 --> (C) H2O + (D) CO2 The first thing I always look for is an element that is in just one reactant and one product. In this case, Hydrogen comes only from glucose and ends up only in water. Because we have 12 hydrogen atoms in glucose and 2 in water, we know there must be 6 times as many water molecules as glucose molecules. Thus, 6C=A We can also see that Carbon atoms come only from glucose and end up only in carbon dioxide. Because there are 6 times as many atoms in one molecule of glucose as there are in one molecule of carbon dioxide, we also know that 6D=A Since 6C=A=6D we know that 6C=6D which means C=D We still haven't balanced oxygen, but let's just try putting in 1 for C and D and 6 for A. 1 C6H12O6 + (B) O2 --> 6 H2O + 6 CO2 We are trying to find B. On the left side of the equation, 6 atoms of oxygen come from glucose. On the right side of the equation, 6 atoms of oxygen are in water and 12 are in carbon dioxide. Thus, we know that the left side needs a total of 12 more oxygen atoms to be balanced. Thus, there must be 6 Oxygen molecules to balance the equation. Our final balanced equation is then: C6H12O6 + 6 O2 --> 6 H2O + 6 CO2 To check our answer we just total up the atoms on each side. ----------------------------------------------------------------- This was the thought process that I would go through to balance this equation. I hope this helps a bit. If you ever get stuck, just try guessing a number and see if it works. Here is another example that is more difficult: CH2S + F2 CF4 + HF + SF6 Again I will name the coefficients: (A) CH2S + (B) F2 (C) CF4 + (D) HF + (E) SF6 Let's assume there is one molecule of CH2S on the left side: 1 CH2S + (B) F2 (C) CF4 + (D) HF + (E) SF6 Looking at the right side, we see that the only compound that contains carbon is CF4. Thus, there must also be 1 molecule of CF4: 1 CH2S + (B) F2 1 CF4 + (D) HF + (E) SF6 Like Carbon, there is only one compound that contains hydrogen on the right side, HF. Because there is 1 atom of Hydrogen in HF and 2 in CH2S, we know there must be twice as much HF as CH2S. 1 CH2S + (B) F2 1 CF4 + 2 HF + (E) SF6 Again, we have one atom of Sulfur on the left, so there must be one also on the right. The only compound that contains sulfur on the right is SF6. We must therefore have one molecule of SF6 to balance the sulfur. 1 CH2S + (B) F2 1 CF4 + 2 HF + 1 SF6 Now the only thing left to balance is Fluorine. There are zero fluorine atoms in the compound CH2S. Looking at the right side, there are 4 in CF4, 1 in each molecule of HF and 6 in SF6. Thus we have a total of 4+2*1+6=12. Thus, we must add 12 fluorine to the left side. Because there are two atoms of fluorine in each molecule of F2, we must add 6 molecules of F2 to the left side. Our final balanced equation is: 1 CH2S + 6 F2 1 CF4 + 2 HF + 1 SF6 I hope this helps. If you ever get stuck, just guess a number for a single compound and then balance the others based on that.