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NISA [10]
3 years ago
6

what volume of carbon 4 oxide measured at STP will produce when 21 gram of sodium hydrogen carbonate for is completely decompose

d accor. ding to the equation. (Na=23. H=1 C=12 O=16) molar vol =22.4dmcube. pleeeeeeeeeeese help​,​
Chemistry
1 answer:
Arlecino [84]3 years ago
8 0

The volume of CO₂ at STP = 2.8 L

<h3>Further explanation</h3>

Given

21 gram of sodium hydrogen carbonate-NaHCO₃

Required

Volume of CO₂

Solution

The decomposition of sodium bicarbonate into sodium carbonate, carbon dioxide, and water :

<em> 2 NaHCO₃(s) → Na₂CO₃(s) + CO₂(g) + H₂O(g)</em>

mol of NaHCO₃ :

= mass : MW NaHCO₃

= 21 g : 84 g/mol

= 0.25

From the equation, mol ratio of NaHCO₃(s) :CO₂(g) = 2 : 1, so mol CO₂ :

= 1/2 x mol NaHCO₃

= 1/2 x 0.25

= 0.125

At STP, 1 mol gas = 22.4 L, so for 0.125 mol :

= 0.125 x 22.4 L

= 2.8 L

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3 years ago
3. A compound was analyzed and found to contain 9.8 g of nitrogen, 0.70 g of
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Answer:

HNO₃

Explanation:

Data given

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Hydrogen =  0.70 g

Oxygen = 33.6 g

Empirical formula = ?

Solution:

Convert the masses to moles

For Nitrogen

Molar mass of N = 14 g/mol

                              no. of mole = mass in g / molar mass

Put value in above formula

                          no. of mole = 9.8 g/ 14 g/mol

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                           mole of N = 0.7 mol

For Hydrogen

Molar mass of H = 1 g/mol

                     no. of mole = mass in g / molar mass

Put value in above formula

                     no. of mole = 0.70 g/ 1 g/mol

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For Oxygen

Molar mass of O = 16 g/mol

                       no. of mole = mass in g / molar mass

Put value in above formula

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mole of O = 2.1 mol

Now we have values in moles as below

N = 0.7

H = 0.7

O = 2.1

Divide the all values on the smallest values to get whole number ratio

N = 0.7 / 0.7 = 1

H = 0.7 / 0.7 = 1

O = 2.1 / 0.7 = 3

So all have following values

N = 1

H = 1

O = 3

So the empirical formula will be HNO₃ i.e. all three atoms in simplest small ratio.

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Answer:

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Explanation:

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