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4 years ago

Nahco3 is the active ingredient in baking soda. how many grams of oxygen are in 0.72 g of nahco3?

1 answer:

4 0

M (O) = M x n

M (O) = 16g/mol

n (O) = 3 x n (NaHCO3)

n (NaHCO3) = m/M

m (NaHCO3) = 0.72g

M (NaHCO3) = 23 + 1 + 14 + 16 x 3 = 86g/mol

n (NaHCO3) = 0.72/86 = 0.008372mol

n (O) = 3 x 0.008372 = 0.02512

m (O) = 16 x 0.02512 = 0.402g

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3 years ago

Carbon tetrachloride reacts at high temperatures with oxygen to produce two toxic gases, phosgene and chlorine. CCl4(g) + (1/2)O

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The question is incomplete, here is the complete question:

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$CCl_4(g)+\frac{1}{2}O_2(g)\rightleftharpoons COCl_2(g)+Cl_2(g);K_c=4.4\times 10^9$ at 1,000 K

Calculate Kc for the reaction $2CCl_4(g)+O_2(g)\rightleftharpoons 2COCl_2(g)+2Cl_2(g)$

Answer: The value of $K_c'$ for the final reaction is $1.936\times 10^{19}$

Explanation:

The given chemical equations follows:

$CCl_4(g)+\frac{1}{2}O_2(g)\rightleftharpoons COCl_2(g)+Cl_2(g);K_c$

We need to calculate the equilibrium constant for the equation, which is:

$2CCl_4(g)+O_2(g)\rightleftharpoons 2COCl_2(g)+2Cl_2(g)$

As, the final reaction is the twice of the initial equation. So, the equilibrium constant for the final reaction will be the square of the initial equilibrium constant.

The value of equilibrium constant for net reaction is:

$K_c'=(K_c)^2$

We are given:

$K_c=4.4\times 10^9$

Putting values in above equation, we get:

$K_c'=(4.4\times 10^9)^2=1.936\times 10^{19}$

Hence, the value of $K_c'$ for the final reaction is $1.936\times 10^{19}$

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3 years ago