5C(s)+2SO2 yields CS2(l)+4CO(g) so how many moles of CS2 form when 2.7 mol of C reacts?
374254
1 mole C gives 0.2 mole CS2\n2.7 mole C gives 0.2 X 2.7 MOLE CS2\n                =0.54 molCS2
849074
K2S + COCl2 --> 2KCl + COS
394069
3C2H5OH + 2K2Cr2O7 + 8H2SO4=3CH3COOH + 2K2SO4 +2Cr2(SO4)3\n                                        +11H2O\n\nFirst you must calculate the (average)oxidation number of central atom of main compounds in reactants.\nEx;Cr oxidation no=+6 in K2Cr2O7\n   C  oxidation no=-2 in C2H5OH\nThen in the products.\nEx;Cr oxidation no=+3 in Cr2(SO4)3\n   C  oxidation no=0 in CH3COOH\n\nThen you must evalute how many electrons must be transferred.\nEx;Oxidation no of Cr is changing from +6 to +3 \n   So it is reduced.\n   SO each atom of Cr must receive 3 electrons.\n   So totally 6 electrons must be received by K2Cr2O7\n   \n   Oxidation no of C is changing from -2 to 0\n   SO each atom of C must donate 2 electrons.\n   So totally 4 electrons must be donated by C2H5OH\n\nThen you must write now a explantion reaction.\nEx;you must write oppsite (simple)propotion of electrons in front of each compound\n  3C2H5OH + 2K2Cr2O7\n \nThen you can calculate how many Cr,C(,K) atoms should be there\n So, 3C2H5OH + 2K2Cr2O7=3CH3COOH + 2K2SO4 +2Cr2(SO4)3\n\nThen you can calculate the medium(SO42-) should be there\n i think 8 SO4 should be there\n\nSo,3C2H5OH + 2K2Cr2O7 + 8H2SO4=3CH3COOH + 2K2SO4 +2Cr2(SO4)3\n\nFinally you must equilize H,O atoms(balanced by water)\nSo,3C2H5OH + 2K2Cr2O7 + 8H2SO4=3CH3COOH + 2K2SO4 +2Cr2(SO4)3\n                                        +11H2O\n\nSecond one ................\nC2H5OH+4I2+6NaOH=CHI3+HCOONa\n                  +5NaI+5H2O
911009
C6H6O + 7O2 yields 3H2O + 6CO2
986414
Start by balancing the equation:\n2Ca3(PO4)2 (s) + 6H2SO4 (aq)-----> 6CaSO4 (s) + 4H3PO4(aq)\n\nThen figure out the molar masses of each of the reactants:\nCa3(PO4)2 = 3(40.078) + 2(30.973762) + 8(15.9994) = 310.176724\n\nH2SO4 = 2(1.00794) + 32.065 + 4(15.9994) = 98.07848\n\nDivide each of your known masses of reactants by its respective molar mass to determine how many moles of each you have, and divide this number by the coefficient (the number in front) of the reactant to determine how many "parts" of each you have:\n(1000/310.176724)/2 = 1.612 parts 2Ca3(PO4)2\n(980/98.07848)/6 = 1.665 parts 6H2SO4\n\nTake the smaller number (since once you run out of one reactant, the reaction is over) and figure out how many moles of calcium sulfate and phosphoric acid you'll produce:\n(1.612)(6)(40.078 + 32.065 + 4(15.9994)) = 1316.7518832g CaSO4\n\n(1.612)(4)(3(1.00794) + 30.973762 + 4(15.9994)) = 631.872933536g H3PO4\n\nSince the total mass of reactants and products needs to be the same, you can check your (er, my) work by totalling up both sides:\n(1.612)(2)(310.176724) + (1.612)(6)(98.07848) = 1948.625g\n1316.7518832 + 631.872933536 = 1948.625g
605005
NaHCO3(s) + CH3COOH(aq) → CH3COONa(aq) + CO2(g) + H2O(l) \nThis is the real chemical reaction.
887797
It looks like you got it. This is how it came out for me:\n\nThis is the reaction:\nC10H8(s) + 12 O2(g) ® 10 CO2(g) + 4 H2O(l)\n\nand Hc=(Summation of Hf of products)-(Summation of Hf of reactants)\nSo...\n\nHc (C10H8) = 10 Hf(CO2) + 4 Hf(H2O) – Hf(C10H8)\n-5156.8kJ/mol=(10)(-393.5 kJ/mol)+(4)(-285.9kJ/mol0-Hf (C10H8)\n-5156.8kJ/mol=(-3935 kJ/mol)+(-1143.6 kJ/mol)- Hf (C10H8)\n-5156.8kJ/mol=(-5078.6)- Hf(C10H8)\n-78.2kJ/mol=-Hf (C10H8)\n78.2kJ/mol=Hf (C10H8)\n\nI believe that's how it's done, but it's been a while since I've done that kind of problem, so it's probably not a terrible idea to double check with someone else.
57069
Balancing we get, S8 + 4Cl2 ---> 4S2Cl2\n\nNo. of Moles = Weight/Molecular Mass \nThus, Moles of sulfur   = 56.68/(8 x 32) = 0.22\n      Moles of chlorine = 97.05/(2*35.5) = 1.37\n\nWe know 1 mole of sulfur requires 4 moles of Chlorine (Cl2)\nTherefore 0.22 moles sulfur will require 0.88 moles of Cl2\nThus sulphur being in limited amount will decide the amount of product formed i.e. it is limiting reagent.\n\nSince 1 mole sulfur produces 4 mole S2Cl2 thus 0.22 moles sulfur will produce 0.88 moles S2Cl2. Therefore weight of S2Cl2 produced is = moles x Molecular mass (S2Cl2)\n                  = 0.88 x (2*32(S2) + 2*35.5(Cl2))\n                  = 0.88 x 134\n                  = 117.92 gm
956142
2Rb(s)+Cl2(g)--2RbCl(s)
132050
oh shit.ok\nn2h2+(ch3)2n2h2+3n2o4 = 5n2 + 6co2 + 5h2
881597
Cr2(SO4)3 + 6KOH  --> 2Cr(OH)3 + 3K2SO4
747676
2C12H22O11 +24O2 ---> 24CO2 +22H2O\na)F\nb)F\nc)T\nd)F\ne)T\nf)T
808775
Na2C2O4\n Na +1, C +3, O -2
844628
a) 53.4\n34gCO  x 1mol CO/28gCO x 2molCO2/2molCO x44gCO2/1molCO2 = 53.4gCO2\nThe 2 moles CO cancel the 2 moles CO2 (from the balanced equation).
855199
7.62 moles of CCl4 containes 4*7.62=30.46 moles of Cl\n30.46*6.023*10^23=1.836*10^25 Cl atoms
382458
H2 + Cl2 -> 2HCl\n2Ag + H2S -> Ag2S + H2 \nCH4 + 2O2 -> CO2 + 2H2O\n2HgO -> 2Hg + O2\n3Fe + 4H2O -> Fe3O4 + 4H2
455269
H2SO4 + Mg(OH)2 -> 2H2O + MgSO4