A sample of 5.58g of Mg(OH)2 is added to 25.4mL of 0.200 M HNO3. Part A Write th

ID: 837676 • Letter: A

Question

A sample of 5.58g of Mg(OH)2 is added to 25.4mL of 0.200 M HNO3.

Part A

Write the chemical equation for the reaction that occurs.

Express your answer as a balanced chemical equation. Identify all of the phases in your answer.

Part B

Which is the limiting reactant in the reaction?

Express your answer as a chemical formula

Part C

How many moles of Mg(OH)2 are present after the reaction is complete?

Part D

How many moles of HNO3 are present after the reaction is complete?

Part E

How many moles of Mg(NO3)2 are present after the reaction is complete?

the balanced reaction is

2HN03 (l) + Mg(0H)2 (s) ----> Mg(N03)2 (s) + 2H20 (l)

given 5.58 g of Mg(0H)2

moles of Mg(0H)2 = 5.58 / 58

moles of Mg(OH)2 =0.0962

moles of HN03 = 25.4 x 0.2 /1000

moles of HN03 = 0.00508

from the reaction

moles of Mg(OH)2 required = moles of (HN03)/2

moles of Mg(0H)2 required = 0.00508 /2

moles of Mg(OH)2 required = 0.00254

so here

Mg(OH)2 is excess and HN03 is the limiting reagent

C) moles of Mg(OH)2 present = 0.0962 - 0.00254

moles of Mg(0H)2 present = 0.09366

D) as HN03 is the limiting reagent

moles of HN03 present = zero

E) moles of Mg( N03)2 = moles of Mg(OH)2 reacted

moles of Mg(NO3)2 present = 0.00254

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