Balance the following equations. e. SnCl6^2- + Fe(CN)6^4- = SnCl4^2- + Fe(CN)6^3

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1) Looks like the N is easiest to start with. There are 2N on each side. Thus, there must be the same (NH4)2 and N2. There are 8H on the left and 2H on the right. So there must be 4 times more H2O than (NH4)2. That would result in 4O + 3O = 7O on the right, and there are already 7O on the left. Finally, check the Cr. There is already only one on each side. (NH4)2 + Cr2O7 --> Cr2O3 + N2 + 4H2O 2) There is one N on the left and two on the right, so there must be twice as much NH3 as N2. So there will be 6H on the left, and we will need 3 H2O to balance that. Now there is 3O on the right, so we need 3 CuO on the left to balance it. Finally, We would need 3Cu on the right to balance with the CuO on the left. 2NH3 + 3CuO --> N2 + 3Cu + 3H2O 3) Its easiest to start with the F. There are 6 on the left, so we need 6 NaF on the right to balance it. Now we have 6 Na on the right. Leave the Na2SiF6 alone. That contains 2 Na, so multiply the lone Na molecule by 4 to balance it. Finally, there is already one Si molecule on each side, so that is balanced. Na2SiF6 + 4Na --> Si + 6NaF 4) Start with the C. There is 4 on the left, so CO2 must be 4 times larger than C4H10. There are 10 H in C4H10, and only 2 in H2O so H2O must be 5 times larger than C4H10. Now balance the O. 4 CO2 would yield 8 O molecules and 5 H2O would yield 5 more, for a total of 13 O on the right side. There are 2 on the left side, which is not divisible into 13. Thus, we need to multiply all of our previous quantities by 2 (2 C4H10, 8 CO2, and 10 H2O). This would yield 26 O on the right side, so we need 13 O2 on the left to balance it. 2C4H10 + 13O2 --> 8CO2 + 10H2O

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