1 g of a mixture of NaHCO3 and Na2CO3 is heated to 150° C . The volume of the CO2 produced at STP is 112.0 mL. Calculate the percentage of Na2CO3 in the mixture (Na = 23, C=12,O =16)

2NaHCO3⟶150∘CNa2CO3+CO2+H2OnNaHCO3nCO2=21nNaHCO3=2nCO2=2×11222400=0.01moleWNaHCO3=0.01×84=0.84gWNa2CO3=1.00−0.84=0.16g%Na2CO3=16

1 g of a mixture of NaHCO3 and Na2CO3 is heated to 150° C . The volume of the CO2 produced at STP is 112.0 mL. Calculate the percentage of Na2CO3 in the mixture (Na = 23, C=12,O =16)

2NaHCO3⟶150∘CNa2CO3+CO2+H2OnNaHCO3nCO2=21nNaHCO3=2nCO2=2×11222400=0.01moleWNaHCO3=0.01×84=0.84gWNa2CO3=1.00−0.84=0.16g%Na2CO3=16

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1 g of a mixture of NaHCO₃ and Na₂CO₃ is heated to 150° C . The volume of the CO₂ produced at STP is 112.0 mL. Calculate the percentage of Na₂CO₃ in the mixture (Na = 23, C=12,O =16)

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Detailed Solution

$2 {NaHCO}_{3} \overset{150^{\circ} C}{⟶} {Na}_{2} {CO}_{3} + {CO}_{2} + H_{2} O$
$\frac{n_{{NaHCO}_{3}}}{n_{{CO}_{2}}} = \frac{2}{1}$
$n_{{NaHCO}_{3}} = 2 n_{{CO}_{2}}$
$= 2 \times \frac{112}{22400} = 0 .01 mole$
$W_{{NaHCO}_{3}} = 0 .01 \times 84 = 0 .84 g$
$W_{{Na}_{2} {CO}_{3}} = 1 .00 - 0 .84 = 0 .16 g$
$% {Na}_{2} {CO}_{3} = 16$