Balance the equation; Ca(OH)2(aq) + Na2Co3(s) = CaCO 3(s) + 2NaOH(aq) How many moles are there in 25.0g of sodium carbonate?
ii) Calculate th mass of sodium hydroxide the scientist could expect to form if he reacts 25.0 of sodium carbonate with an excess of calcium hydroxide.
B) Scientist only found out he made 10.4 of sodium hydroxide. Calculate the percentage yield of this
Balance the equation; Ca(OH)2(aq) + Na2Co3(s) = CaCO 3(s) + 2NaOH(aq) How many moles are there in 25.0g of sodium carbonate?
ii) Calculate th mass of sodium hydroxide the scientist could expect to form if he reacts 25.0 of sodium carbonate with an excess of calcium hydroxide.
B) Scientist only found out he made 10.4 of sodium hydroxide. Calculate the percentage yield of this
So basically you know that mass/molar mass gives you moles so the molar mass of Na2Co3 is 106 and when you do mass/molar mass you should get 0.236 ii) this is kinda poorly worded/set out because if its a solution then I don't know why their talking about it in terms of mass but because they have given you the mass, just work out that and if you look at the ratios using stoichiometry, you should multiply it by 2 to get the moles of NaOH due to 1:2 ratio and then using mass/mm again rearrange the equation to give you mass of NaOH. (Just multiply the moles from stoichiometry with the molar mass of NaOH which was 40) iii) use your mass as the expected yield and work out the % yield. Would be nice if you could list answers next time
Ca(OH)2(aq) + Na2Co3(s) = CaCO 3(s) + 2NaOH(aq) How many moles are there in 25.0g of sodium carbonate?
n(Na2Co3)=25/(46+12+48)=0.235849056 mol
ii) Calculate th mass of sodium hydroxide the scientist could expect to form if he reacts 25.0 of sodium carbonate with an excess of calcium hydroxide.
m(NaOH)=n*rfm=(2*0.235849056)*(23+16+1)=18.86g
B) Scientist only found out he made 10.4 of sodium hydroxide. Calculate the percentage yield of this