2307 gkeep the answer with correct sig figshope it is clear..good luck with the exam tomorrow-This is the easy kind to do and youve got about 1/3 of the procedure.1)Balanced equation. Ca(NO3)2 + Na2CO3 -> CaCO3 + 2 NaNO32)Convert the mass into moles0.;75 g Are you sure it is that and not 75.......
=.7949 g of Na2CO3
similarly you can calculate the calcium nitrate amount also. it is again 1:1 mole ratio there for amount in moles is .0075 and mass is .0075 * 164.09 g = 1.2307 g
keep the answer with correct sig figs
hope it is clear..
good luck with the exam tomorrow
This is the easy kind to do and you've got about 1/3 of the procedure.
1) Balanced equation. Ca(NO3)2 + Na2CO3 -> CaCO3 + 2 NaNO3
2) Convert the mass into moles
0.;75 g Are you sure it is that and not 75.0g? Well, just follow the procedure
0.75 g of CaCO3 = 0.75g / 100.09 g/mol = 0.00749 moles
3) Use the mole ratio between CaCO3 and Na2CO3 to calculate the number of moles of Na2CO3 needed. mole ratio is 1 to 2, or for every mole of CaCO3 produced you only need 1/2 that many moles of Na2CO3. So moles of Na2CO3 needed is 0.00749 / 2 = 0.00375 moles
4) Convert moles back to mass
mass of Na2CO3 = 0.00375 moles x 105.99 g/mol = 0.397 g
5) Do the same calculation with Ca(NO3)2 to find the mass of that needed.
Ok so .75g CaCO3 (1mol CaCO3/100g CaCO3) (1mol Ca(NO3)2/1 mol CaCO3) (164 g Ca(NO3)2/1 mol Ca(NO3)2. And then take out the Ca(NO3)2 and plug in the Sodium Carbonate. Sorry if im wrong bro i havent done stoich for a while and im kinda off but im pretty surer its right
