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IrinaVladis [17]
3 years ago
8

a 3.81 g sample of NaHCO3 was completely decomposed. After decomposition, Na2CO3 had mass of 2.86g. Determine mass of H2CO3 prod

uced.
Chemistry
1 answer:
qaws [65]3 years ago
8 0

Answer:

1.67g H2CO3 are produced

Explanation:

Based on the reaction:

2NaHCO3 → Na2CO3 + H2CO3

<em>2 moles of NaHCO3 produce 1 mole of Na2CO3 and 1 mole of H2CO3</em>

To solve this question we need to find the moles of Na2CO3 = Moles of H2CO3. With their moles we can find the mass of H2CO3 as follows:

<em>Moles Na2CO3 -Molar mass: 105.99g/mol-</em>

2.86g Na2CO3 * (1mol/105.99g) = 0.02698 moles Na2CO3 = Moles H2CO3

<em>Mass H2CO3 -Molar mass: 62.03g/mol-</em>

0.02698 moles  * (62.03g/mol) =

<h3>1.67g H2CO3 are produced</h3>

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A reaction of importance in the formation of smog is that between ozone and nitrogen monoxide described by O 3 ( g ) + NO ( g )
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(b) 2.56 mol.L⁻¹.h⁻¹

Explanation:

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The reaction rate of equation (1) is given by:

rate = k*[O_{3}][NO]     (2)

<u>We have:</u>

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Hence, to find the inital reacion rate we will use equation (2):

rate = k*[O_{3}]_{0}[NO]_{0} = 3.91 \cdot 10^{6} M^{-1}s^{-1}*2.35\cdot 10^{-6} M*7.74 \cdot 10^{-5} M = 7.11 \cdot 10^{-4} M/s  

Therefore, the inital reaction rate is 7.11x10⁻⁴ M/s

(b) The number of moles of NO₂(g) produced per hour per liter of air is:

t = 1 h

V = 1 L

\frac{\Delta[NO_{2}]}{\Delta t} = rate

\frac{\Delta[NO_{2}]}{\Delta t} = 7.11 \cdot 10^{-4} M/s*\frac{3600 s}{1 h} = 2.56 mol.L^{-1}.h{-1}

Hence, the number of moles of NO₂(g) produced per hour per liter of air is 2.56 mol.L⁻¹.h⁻¹

I hope it helps you!                                

5 0
3 years ago
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