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Vikentia [17]
2 years ago
9

For the reaction HNO3 + Mg(OH)2→ Mg(NO3)2 + H2O, how many grams of magnesium nitrate are produced from 8.00 mol of nitric acid,

HNO3? Balance the equation
Chemistry
2 answers:
ollegr [7]2 years ago
6 0

Answer:

tha mass of magnesium nitrate is 592g

Explanation:

from a balanced chemical equation

2HNO3 + Mg(OH)2→ Mg(NO3)2 + 2H2O

2 mol of nitric acid is equivalent to 1 mol of magnesium nitrate. then 8 mol of nitric acid will be equivalent to 4 mol of magnesium nitrate.

oee [108]2 years ago
5 0

Answer:

2HNO3 + Mg(OH)2 --> 2H2O + Mg(NO3)2

593.2 grams Mg(NO3)2

Explanation:

First, balance the equation

The first thing I notice is that there are 2 nitrates on the reactant side so I would put a 2 in front of HNO3. Now the problem is that there are 4 hydrogens on the reactants side, so add a 2 to H2O.

Next is stoichiometry

We are looking for Mg(NO3)2 so use the balanced equation to create a mole ratio and then convert to grams.

8 mol HNO3 * 1 mol Mg(NO3)2/2 mol HNO3 * 148.3 g Mg(NO3)2/1 mol Mg(NO3)2 = 593.2 grams Mg(NO3)2

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1. 0.02 M

2. 0.01 M

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1.

Concentration of HCl = 0.05 M

end point comes at = 10 ml

So, concentration of OH⁻(aq) = [OH⁻(aq)] ⇒ (0.05 × 10) ÷ 25 ⇒ 0.02 M

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3.

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K_{sp} = [0.01 × (0.02)²] = 4×10⁻⁶

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