A sample of 8.00 g of Mg(OH)2 is added to 24.2 mL of 0.205 M HNO3. How many moles of Mg(OH)2 our present after the reaction is c

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1 year ago

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omplete?

1 answer:

kogti [31] · Answer 1 · 2023-05-08T23:49:43+03:00

The amount of Mg(OH)2 present after the reaction is complete is 0.136 moles of Mg(OH)2.

The equation of the reaction is;

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

Number of moles of Mg(OH)2 = 8.00 g/58 g/mol = 0.138 moles

Number of moles of HNO3 = 0.205 M × 24.2 mL/1000 = 0.00496 moles

Given that;

2 moles of HNO3 reacts with 1 mole of Mg(OH)2

0.00496 moles of HNO3 reacts with 0.00496 moles × 1 mole /2 moles = 0.00248 moles of Mg(OH)2

Hence, Mg(OH)2 is the reactant in excess.

The amount of Mg(OH)2 remaining = Amount present - Amount reacted

Hence; 0.138 moles - 0.00248 moles = 0.136 moles of Mg(OH)2