Mass of sodium in 53g of Na2CO3 is?
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in these type of question...we use stocheometry [tex]Mw of Na_2CO_3 = 106 \frac{gram}{mole}
4Na + 2C + 3O_2 --\ \textgreater \ 2Na_2CO_3
at t=equilibrium n(mole)= \frac{wt}{MW} = 53/106 = 0.5 mole
if mole of 2Na_2CO_3 is 0.5 then by stochiometry Na has 1 mole
so for 1 mole Na = \frac{wt}{MW}
1 × 23 = wt
[/tex]
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