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Ad libitum [116K]
1 year ago
7

How many grams of nitric acid HNO₃, are required to neutralize (completely react with) 4.30 grams of Ca(OH)2 according to the ac

id-base reaction: 2HNO₃(aq) + Ca(OH)₂(aq) → 2H₂O(l) + Ca(NO₃)₂(aq) Answer as equation
Chemistry
1 answer:
Brrunno [24]1 year ago
4 0

Answer:

7.32g of HNO3 are required.

Explanation:

1st) From the balanced reaction we know that 2 moles of HNO3 react with 1 mole of Ca(OH)2 to produce 2 moles of H2O and 1 mole of Ca(NO3)2.

From this, we find that the relation between HNO3 and Ca(OH)2 is that 2 moles of HNO3 react with 1 mole of Ca(OH)2.

2nd) This is the order of the relations that we have to use in the equation to calculate the grams of nitric acid:

• starting with the 4.30 grams of Ca(OH)2.

,

• using the molar mass of Ca(OH)2 (74g/mol).

,

• relation of the 2 moles of HNO3 that react with 1 mole of Ca(OH)2 .

,

• using the molar mass of HNO3 (63.02g/mol).

4.30g\text{ Ca\lparen OH\rparen}_2*\frac{1\text{ mol Ca\lparen OH\rparen}_2}{74g\text{ Ca\lparen OH\rparen}_2}*\frac{2\text{ moles HNO}_3}{1\text{ mole Ca\lparen OH\rparen}_2}*\frac{63.02g\text{ HNO}_3}{1\text{ mole HNO}_3}=7.32g\text{ HNO}_3

So, 7.32g of HNO3 are required.

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Answer:

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For 2.5 moles we multiply the above chemical eq. With 2.5

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3 years ago
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3 years ago
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Answer:

1.14 angstroms

Explanation:

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