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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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Answer:

MnSO₄.7H₂O

Explanation:

To solve this question, we need to convert the mass of the dehydrated MnSO₄. The difference between mass of the hydrate and dehydrated compound is the mass of water. With the mass we can find the moles of water and the formula of the hydrate:

<em>Moles MnSO₄ -Molar mass: 151g/mol-:</em>

17.51g * (1mol / 151g) = 0.116 moles

<em>Moles H₂O -Molar mass: 18g/mol-:</em>

32.14g-17.51g = 14.63g * (1mol / 18g) = 0.813 moles

The ratio of moles MnSO₄: Moles H₂O represent the amount of water molecules in the hydrate:

0.813mol / 0.116mol = 7 molecules of water.

The hydrate formula is:

<h3>MnSO₄.7H₂O</h3>
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