171k views
3 votes
For the reaction 2hno3 + mg(oh)2 → mg(no3)2 + 2h2o, how many moles of magnesium nitrate are produced from 8.00 moles of nitric acid, hno3?

2 Answers

4 votes

Final answer:

To find out how many moles of magnesium nitrate (Mg(NO3)2) are produced from 8.00 moles of nitric acid (HNO3), we need to use the balanced chemical equation provided. The balanced equation is: 2HNO3 + Mg(OH)2 -> Mg(NO3)2 + 2H2O. From the equation, we can see that 2 moles of HNO3 react with 1 mole of Mg(NO3)2. Therefore, if we have 8.00 moles of HNO3, we would produce half as many moles of Mg(NO3)2, which is 4.00 moles.

Step-by-step explanation:

To find out how many moles of magnesium nitrate (Mg(NO3)2) are produced from 8.00 moles of nitric acid (HNO3), we need to use the balanced chemical equation provided. The balanced equation is: 2HNO3 + Mg(OH)2 -> Mg(NO3)2 + 2H2O. From the equation, we can see that 2 moles of HNO3 react with 1 mole of Mg(NO3)2. Therefore, if we have 8.00 moles of HNO3, we would produce half as many moles of Mg(NO3)2, which is 4.00 moles.

User Unconditional
by
6.9k points
3 votes
you would get four moles of magnesium nitrate :) you would have to
“ ?molesmg(oh)2 = 8molmg(no3)2 x molmg(oh)2 / 2molhno3 = 4 moles of magnesium nitrate :))) hopefully this helps! <3
User Marcin Erbel
by
6.9k points