write the number of moles of H2O produced in a reaction of 50ml 1.0M HC2H3O2 and 50ml 1.0 M NaOH?
955158
Find the moles of the acid CH3COOH and of the NaOH:\n\nn = C*V = 1*0.05 = 0.05 mol CH3COOH\n\nn = C*V = 1*0.05 = 0.05 mol NaOH\n\nAccording to the balanced chem. equation:\n\nCH3COOH + NaOH  --> CH3COONa + H2O\n\n1 mol of CH3COOH reacts with 1 mol of NaOH and gives 1 mol of H2O\n\nSo 0.05 mol CH3COOH reacts with 0.05 mol NaOH and gives 0.05 mol of H2O\n\nThat's the answer: 0.05 mol H2O
1107016
To find amount of water produced, first , we need to write a balanced equation for the reaction:  \n\n2H2 + O2 = 2H2O\n\nTo do any Stoechiometry calculations we need to change the grams to moles.\n\n1 g H2 * 1 mol/2.016 g = .496 moles of H2\n10 g O2 * 1 mol /32 g =.3125 moles \n\nThe ratio between H2 and H2O in the balance equation is 1:1\nOn the other hand, the ratio between O2 and H2O in the balanced equation is 1:2.  \nThat means we can make .496 moles of H2O from the given H2 and 03125*2 = .625 moles of H2O from the given O2.  In this case H2 is the limiting reactant since H2 will run out first. That means we can only make .496 moles of H2O from the given H2 and O2. \nNow we need to change the moles of H2O to grams.  \n.496 moles * 18.016 g/1 mol H2O = 8.936 grams.
867158
2Mg+O2-->2MgO\n2MgO+2H2o---->2Mg(OH)2
954811
C3H8 + 5O2 --> 3CO2 + 4H2O
318573
The reaction between Hydrochloric acid and Sodium Hydroxide is as follows,\n\nHCl + NaOH --> NaCl + H2O\n\nYou can see that there is a 1:1 ration between the number of moles of HCl needed to completely reacted with NaOH.\n\nSince you are given both a volume and concentration of the NaOH solution, you should be easily able to calculate the number of moles of NaOH availible.\n# mole = volume * concentration\n# moles = .1 Liters * .25 moles/L = .025 moles of NaOH\n\nIf you know you have .025 moles of NaOH, you know you need .025 moles of HCl to react it completely.\n\nIf you know that you have a .15 mole/Liter solution of HCl, you can calculate the volume required to get .025 moles lie this,\nLiters = moles / oncentration\nLiters = .025 moles / .15 moles/Liter = .01667 Liters or about 167 mL of solution.\n\n\nNext time do your own homework.
929350
Do you mean M instead of N?\nIf you mean M, then\n1,00M/L*0,500 L= 0,500M\n0,500 moles NaOH*(40g NaOH/1 mol NaOH)= 20,0g of NaOH\nIt is c).
559540
the formula M=moles/L\n first convert mL into L\n(500)(10^-3) = .5L\nM= .50moles/.5L =  1M
116913
I think it is this:\n\n            \nC12H22O11(s) --- HNO3---->  12C(s) = 11H2O(g) + heat
904934
2HNO3 + Mg(OH)2 -->Mg(NO3)2 +2H20\n\n2H+(aq) + 2NO3-(aq) + Mg+(aq) + 2(OH)-(aq) -->Mg+(aq) + 2(NO3)-(aq) +2H20(l)\n\n2+(aq) + 2(OH)-(aq) --> 2H20(l)
697545
1- find heat quantity\n Q=C*t\nQ = heat quantity of the reaction\nC= heat capacipy of calorimeter\nt=temperature increasing\n\n2- find heat of reaction \n H=(Q*1000)/(N*V)\nN = normality of acid and base\nV= volume of acid and base
295951
2 CH3OH + 3 O2 → 2 CO2 + 4 H2O + 726 kJ/mol
696206
the moles of NaOH = the moles of OH - = 0.125 * 0.606 = 0.07575\nthe moles of OH - in the new solution = 0.07575/(15 + 0.125 ) = 5*10^(-3) M\npOH = - log ( 5*10^ -3) = 2.3\n=> pH = 14 - pOH = 11.7
164045
What do you want a step by step set of instructions or a balanced equation?\n\n2C6H5OH + NaOCL = 2C6H50 + NaCl + H2O
938236
Step one: find the moles of H2SO4.  (2.7 moles per liter)*0.025 liters.\n\nStep two: multiply by two because the reaction is \n2NaOH + H2SO4 ----> 2H2O  + Na2SO4 and so you need two NaOH molecules per molecule of H2SO4.\n\nStep three: multiply by the molecular mass of NaOH.\n\nThat will get you the answer.
697635
1) Calculate the number of moles of NaOH in 300 mL\n3.81 g /( 22.99+16+1.008) = 0.0955 mol \n\n2) Calculate the number of moles of NaOH in the 25 mL transfered\n0.0955 mol * 25/300 = 7.96 x 10^-3 mol\n\n3) Calculate the molarity \n7.96 x 10^-3 mol / 100 mL = 0.0796 mol/liter
474531
CH3COOH + NAOH =.CH3COONA +H2O
868179
2Rb + 2 H20 --> 2RbOH + H2