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2 years ago

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A student heats 10.52 g of sodium hydrogen carbonate in a crucible until the compound completely decomposes to sodium carbonate

(rxn 1 in your lab write-up). Calculate the theoretical yield (in grams) of sodium carbonate for this reaction.

1 answer:

saveliy_v [14]2 years ago

4 0

Answer:

$m_{Na_2CO_3}^{theoretical}=6.636gNa_2CO_3$

Explanation:

Hello!

In this case, since the decomposition of sodium hydrogen carbonate is:

$2NaHCO_3(s)\rightarrow Na_2CO_3(s)+H_2O(g)+CO_2(g)$

Thus, since there is a 2:1 mole ratio between the sodium hydrogen carbonate and sodium carbonate, and the molar masses are 84.01 and 105.99 g/mol respectively, we obtain the following theoretical yield:

$m_{Na_2CO_3}^{theoretical}=10.52gNaHCO_3*\frac{1molNaHCO_3}{84.01gNaHCO_3}*\frac{1molNa_2CO_3}{2molNaHCO_3} *\frac{105.99gNa_2CO_3}{1molNa_2CO_3}\\\\ m_{Na_2CO_3}^{theoretical}=6.636gNa_2CO_3$

Best regards!

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Answer : The [H] is increasing at the rate of 0.36 mol/L.s

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

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From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$2D(g)+3E(g)+F(g)\rightarrow 2G(g)+H(g)$

The expression for rate of reaction :

$\text{Rate of disappearance of }D=-\frac{1}{2}\frac{d[D]}{dt}$

$\text{Rate of disappearance of }E=-\frac{1}{3}\frac{d[E]}{dt}$

$\text{Rate of disappearance of }F=-\frac{d[F]}{dt}$

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$\text{Rate of reaction}=-\frac{1}{2}\frac{d[D]}{dt}=-\frac{1}{3}\frac{d[E]}{dt}=-\frac{d[F]}{dt}=+\frac{1}{2}\frac{d[G]}{dt}=+\frac{d[H]}{dt}$

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As,

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and,

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$+\frac{d[H]}{dt}=0.36mol/L.s$

Thus, the [H] is increasing at the rate of 0.36 mol/L.s

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