How much 4 M KOH is required to react with 3.845g of Al metal?
816883
First find the moles of Al. The relative atomic mass of Al equals Ar = 27, so\n\nn = m/Ar = 3.845/27 = 0.142 mol (approx)\n\nFrom the chem. equation:\n\n2Al + 6KOH  --> 2K3AlO3 + 3H2\n\n2 mol of Al reacts with 6 mol KOH, so\n0.142 mol Al reacts with 6*0.142/2 = 0.427 mol KOH\n\nAt last find the volume of the KOH solution:\n\nn = C*V, V = n/C, V = 0.427/4, V = 0.107 L or 107 mL (approx)
890642
Al + Fe(NO3)3 ---> Al(NO3)3 + Fe.\n\n2K + 2H2O ---> 2KOH + H2\n\n2KOH + CuSO4 ---> Cu(OH)2 + K2SO4\n\n3NaOH + H3PO4 ---> 3H2O + Na3PO4\n\nBest?
895751
Al2S3 + 6H2O -> 2Al(OH)3 + 3H2S\n\nP4O10 + 6H2O -> 4H3PO4\n\n\nShould be right.
894676
Al(CH3COOH)3 means that in each molecule of that salt, you have one aluminium and three acetate.  Each acetate contains two carbon, two oxygen and four hydrogen.  So in total, you have: 1 Al, 6 C, 12 H, 6 O.\n\nBecause you state that you have one mole of that aluminium acetate, this means that you have 1 mole of Al, 6 moles of C, 12 moles of H, 6 moles of O, so 25 moles of atoms total.\n\nIf you had more salt, lets say 4,8 moles of Al(CH3COOH)3, then you would have to multiply everything by 4,8.
897929
convert 995.0 g CH4 into moles and then multiply that answer with 882kJ/mole -- make sure your units cancel
1022888
mass of KOH:\n(.0313L)(0.118M)(56.11g/mol) = 0.21g\n\nYour question probably asked for the concentration of KOH:\n(.0313L)(.118M)/(.0287L) = 0.13M
223710
Start by balancing Al, then Cl. Al is already  balanced, so when you balance Cl everything is easy from their. \nYou final amswer will be:\n3HCl + Al(OH)3 ----> 3H2O + AlCl3
882018
1) 2Al + Fe2O3=Al2O3 + 2Fe\n2) 8.53g BaCl2*(1mole BaCl2/ 208.3G BaCl2)=.0410mole BaCl2\n.0410mole BaCl2*(2moles KCl/1moleBaCl2)=.0820 mole KCl2\n.0820mole KCl2*(110.1gKCl2/1mole KCl2)=9.03gKCl2\n3)i don't know this one\n4) 2NaOH + H2 is reduced\n5) i don't know this one\n6) 5gNa*(1moleNa/23gNa)=.217mole Na\n.217mole Na*(6.02x10^23/1mole Na)=1.31x10^23 atoms Na
290781
MgSO4(s) + 2HCL(aq) ---> H2SO4(aq) + MgCl2 (aq)
963470
CO + 2H2  --->  CH3OH\n\nmoles H2: 2.00g/ 1g/mol = 2.00 mol\nmoles CO: 18.50g / 28g/mol = 0.6607mol\n\nyour limiting reactant is CO since for every 0.6607 moles of CO, you'll only need 1.321 moles H2. there's an excess of 0.6790 moles H2.\n\ntheoretical yield, therefore, is (1.321/2)*100 =  66.05%\n\nbased from the balanced equation, there should also be 0.6607 moles of CH3OH (MW = 32g/mol) that will be formed. so,\n\nweight of CH3OH = 0.6607mol * 32g/mol = 21.14 grams
39171
2Al + 3S --> Al2S3\n\nConvert the grams to moles and use stoichiometry to get the moles of Al2S3.
783725
All right, you know what M means, right?  It's short for moles per liter.  You want to have 0.250 moles of Al2(SO4)3 in each liter of solution.  You're preparing 0.4 L of solution, so you need (0.4 * 0.250) = 0.1 moles of Al(SO4)3.\n\nNow the molar mass of each element is listed on most periodic tables.  That's how much one mole of that atom weighs.  A molecule weighs as much as all the atoms in it put together, so a mole of Al2(SO4)3 weighs as much as two moles of Al, 3 moles of S and 12 moles of O.  That's 2(26.982) + 3(32.06) + 12(15.999), according to my periodic table.  So if you add it all up, a mole of Al2(SO4)3 weighs 342.132 grams.  You only need a tenth of a mole, so you use 34.2132 grams.
843494
Forget the potassium ions - they don't do anything. The aluminium metal dissolves in the potassium hydroxide solution, forming aluminate ions and hydrogen gas.\n\nThe ionic equation is:\n\n2Al + 2OH- + 6H2O ---> 2[Al(OH)4]- + 3H2\n\nTo explain how to get this equation, you need the ideas of half equations, and oxidation and reduction.\n\nFor example, Al + 4OH- ---> [Al(OH)4]- + 3e-\n\nand 4H2O + 4e- ----> 2H2 + 4OH-\n\nNow combine them, getting the number of electrons equal, and cancelling down some hydroxide ions.\n\nIf it's any consolation, this exercise would be beyond most A-level pupils in the UK.
969093
H2 = 2g/mol   2.05g = 1.025mol\nCO = 28g/mol    15.85g = .57mol\n\n2H2 + CO --> CH3OH\n\nmust be reduced to \nH2 + (1/2)CO --> (1/2)CH3OH\n\n1.025mol + (1/2).57mol --> the quantity of the limiting reactant which is .285mol CH3OH which at 32g/mol = 9.12g
781203
2H3PO4(aq) + 3Mn(OH)2(s)  --> Mn3(PO4)2(s) + 6H2O(l)
580085
I think you are using Aluminum Sulfate, which is Al2(SO4)3. Molarity is defined as Moles per Liter, so the first thing to do is convert 210g Al2(SO4)3 to Moles. There are approximately 342.3g in one mole of Al2(SO4)3, so, after converting, you have about .613 Moles of Aluminum Sulfate. Then you can simply divide .613 by 2.75, giving you a molarity of .223.
24431
3225..............x\nZn   +  2HCl -> ZnCl2 + H2\n65.4............136.4\n\nx=3225*136.4/65.4\nx= 6 726.14679 grams