How many moles of sodium carbonate na2co3 are contained by 57.3 g of sodium carbonate
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Answer:
Maximum mass of produces=50.8g;
Explanation:
Firstly balance the chemical reaction,
Molecular weight of =334g/mole;
Molecular weight of CO =28g/mole;
Number of mole of given=80g/(334g/mole)=0.239mole;
Number of mole of CO given=28g/(28g/mole)=1 mole
Moles are:
Number of mole of =0.239
Number of mole of CO=1;
From the balanced equation,
1 mole of reacts with 5 moles of CO;
Hence;
0.239 mole of will react with 5*0.239 moles of CO means
Moles of CO that reacts with =1.2 mole but we have given 1 mole therefore
Will be excess reagent and CO will be the limitting reagent
And mass of product depends on limiting reagent i.e. mass of CO.
5 mole of CO produces 1 mole of
;
1 mole of CO produces 1/5 mole of ;
Maximum mass of produces=mole*molecular weight;
Molecular weight of =254g/mole;
Maximum mass of produces = (1/5)*254=50.8g;
Maximum mass of produces=50.8g;