How many number of moles are present in 3.6gram of water (H2O)?
As we know,Mole = weight in gram / molar massHere, weight = 3.6 gmMolar mass (H2O) = (2*1)+16 = 18By putting the values..Moles = 3.6/18 = 0.2Therefore 0.2 mole are present in 3.6 gms of H20.Hope it will help you..Thanks..:-)
Calculate the max # of moles and grams of H2S that can form when 200. g aluminum sulfide reacts with water?
Calculate the maximum number of moles and grams of H2S that can form when 200. g aluminum sulfide reacts with 150. g water. Al2S3 + 6H2O --> 2Al(OH)3 + 3H2S ____ mol ____ g What mass of the excess reactant remains? ____ g Unsure of how to solve this, I would appreciate some help, thanks.
Can you please help me with these questions?
Terminology: n=number of moles, m=mass in grams, M=molecular mass (from periodic table of elements) To find number of moles: n(Mg)=? m(Mg)=72g M(Mg)=24 n=m/M n(Mg)=m(Mg) / M(Mg) =72 / 24 = 3 moles To find the mass: m(Cl2)=? M(Cl2)= 35.5 x 2 = 71 n(Cl2)= 5 moles m= n x M = 5 x 71 = 355 grams I hope this helps you do the rest. Make sure you set it out clearly like I have.
In order to completely neutralize 20 mL of a solution of HCl 0.1 M, 40 mL of a solution of NaOH must be added. What is the M of the NaOH solution?
No, this working is incorrect. You tripped up on two points.20 milliliters is what decimal fraction of a liter?milli- means one thousandth, so 20 mL = 20/1000 liters20/1000 = 2/100 = 1/50 literDivide one by fifty:50 ) 1 wont go , result 0.? so try50) 10 won’t go result 0.0? so try50)100 goes twice result 0.02So 0.02 L = 20mL Result for VaYou made the same mistake with Vb = 40 mL SHOULD BE = 0.04 LThe second mistake was more important because in this equation the first two scaling mistakes canceled out!You wrote: 1 mol x 0.1 M x 0.20 L / 1 mol x 0.40 L = 8x10-3M1 x 0.1 x 0.2 / 1 x 0.4 = 0.05M NOT 8 millimol !You multiplied (x) instead of dividing ( / )
How many aluminum atoms are in a 270 gram sample of Al (aw =27)?
How many grams of sucrose are in 0.1 moles of sucrose (fw=180)? how many moles of Ca(OH)2 (fw=74) are in 111 grams of Ca(OH)2? balance this equation: Fe(NO3)3 + NaOH ---> Fe(OH)3 + NaOH answer the folowing questions based on the following balanced equation: 2N2 + 3O2 ---> 2N2O3 a) how many moles of oxygen would you need to make 4 moles of N2O3? b) how many moles of N2O3 could you make from 1 moles of O2? c) How many moles of N2 would you need to produce 6 moles of N2O3? based on the following equation answer the following questions: 2H2 + O2 ----> 2H2O a) how many grams of oxygen would you need to make 108 grams of water? b) how many grams of water could be made from 8 grams of hydrogen? in the following equation 56 grams of nitrogen reacted with 16 grams of hydrogen to produce ammonia (NH3). what is the limiting reactant in this reaction N2 + 3H2 ---> 2NH3 which metal is oxidized : CuCl2 + Zn ---> Cu + ZnCl2 what are the spectator ions and the net ionic equation for the following chemical reaction: Pb^2+ (aq) + 2NO3^- (aq) + 2Na^+ (aq) + CrO4^2- (aq) ----> PbCrO4 (s) + 2Na^+ (aq) +2NO3^- (aq)
Chem!! Please help! Calculate the mass of CO2 that can be removed by reaction with 3.88 kg of lithium hydroxid?
Lauren, You need to set up balanced equations for each of the reactions involving carbon dioxide: CO2 + 2LiOH --> H2O + Li2CO3 CO2 + Mg(OH)2 --> H2O + MgCO3 3CO2 + 2Al(OH)3 --> 3H2O + Al2(CO3)2 (a) So, let's figure out the number of moles of 3.88 kg LiOH: moles = (3.88 kg x 1000 g/kg) / (23.95 g/mole LiOH) = 162.00 moles. Since 2 moles of LiOH react with one mole of CO2, moles CO2 = 1/2 x 162.00 = 81.00 moles Therefore, grams CO2 absorbed = 81.00 moles x 44 g/mole CO2 = 3564 gm = 3560 grams (three sig figs). (b) The same procedure can be used for the others, too. Find the moles of each hydroxide compound, convert that to moles of CO2 via the above equations, and turn those into grams CO2: LiOH: [(1.42 g LiOH)/ (23.95 g/mole LiOH)] x [(1 mole CO2) / (2 moles LiOH)] x 44 g/mole CO2 = 1.30 g CO2 Mg(OH)2: [(1.42 g Mg(OH)2)/ (58.33 g/mole Mg(OH)2)] x [(1 mole CO2) / (1 moles Mg(OH)2)] x 44 g/mole CO2 = 1.07 g CO2 Al(OH)3: [(1.42 g Al(OH)3)/ (78.00 g/mole Al(OH)3)] x [(3 mole CO2) / (2 moles Al(OH)3)] x 44 g/mole CO2 = 1.20 g CO2 (three sig figs for each) Hope that helped!