Question

For the reaction 2HNO3 + Mg(OH)2 → Mg(NO3)2 + 2H₂O, how many grams of

magnesium nitrate are produced from 2 grams of nitric acid, HNO3 ?
Round your answer to the nearest tenth. Do not round your answers until the last step of
the problem. If you do, the computer might mark your answer incorrect and you will be
asked to complete your second trial for this assessment.
Use the following molar masses:
Element Molar Mass
Hydrogen 1
Magnesium 24
Nitrogen 14
Oxygen 16.

Answer 1

Step-by-step explanation:

First, we need to determine which reactant is limiting. This can be done by calculating the moles of each reactant present:

2 grams HNO3 * (1 mole HNO3/63.01 grams HNO3) = 0.0317 moles HNO3

2 grams Mg(OH)2 * (1 mole Mg(OH)2/58.33 grams Mg(OH)2) = 0.0343 moles Mg(OH)2

Since there are fewer moles of HNO3, it is the limiting reactant. Using the balanced chemical equation, we can determine the amount of magnesium nitrate produced:

2 moles HNO3 * (1 mole Mg(NO3)2/2 moles HNO3) * (148.31 grams Mg(NO3)2/1 mole Mg(NO3)2) = 148.31 grams Mg(NO3)2

Therefore, 2 grams of nitric acid produces 2/63.01 moles of HNO3 which produces 2/63.01 x 148.31 = 4.43 grams of magnesium nitrate. Rounded to the nearest tenth, the answer is 4.4 grams.