How many grams of NaOH can be formed when 43.0 grams of Na2O react with 27.0 grams of H2O? (1 point) Na2O + H2O yields NaOH Be sure to balance the equation. 13.9 grams 18.0 grams 27.0 grams 55.5 grams How many grams of H2 are needed to react completely with 44.1 grams of C? (1 point) C + H2 yields CH4 Be sure to balance the equation. 7.40 grams 14.7 grams 88.2 grams 132 grams
So lets start with the first one and balance it, are you able to do that?
yes Na2O + H2O = 2 NaOH
Perfect now we need to determine which of the reactants is the limiting reagent to do this we take the mass and divide by the solar mass: 43g Na2O * (1mol Na2O/61.98g/mol)=? 27g H2O * (1mol H2O/18.02g/mol)=?
um..having trouble with that part
Are you able to elaborate which part you are having problems with?
43g Na2O * (1mol Na2O/61.98g/mol)=? 27g H2O * (1mol H2O/18.02g/mol)=? :)
Ok so the first one there we will get 0.694mol of Na2O and 1.50mol of H20 so the limiting reagent will be the Na2O
Knowing that the Na2O is the limiting reagent we can use this to determine the yield of NaOH so: 0.694mol Na2O * (2mol NaOH/1mol Na2O) * (39.99g/mol /1mol NaOH) = ? g NaOH
I think i got18.0 grams
wait idk :/
Sorry I messed something up one sec
Wait no I was looking at the wrong options for a sec lol when I did the math I got the 55.5
ohhh okay I think i messed up with a number too :) but 55.5 i agree too
Awesome so lets do the second one, do you want to balance it?
C + 2 H2 = CH4
Perfect! So for this one we don't need to find the limiting reagent. We can start with: 44.1g C * (1mol C/ 12.01g/mol C) * (2mol H2/ 1molC) * (2.02g/mol H2/ 1mol H2)= ?gH2
You missed multiplying it by another two somewhere in there I believe
I might have multipled by the wrong thing..132 grams i believe
When I went through it I ended with 14.8 since I carried all decimals without decimal carrying I got 14.7
lol i'm sorry i made a simple error hahah :)
That is no problem lol happens to everyone!