Write the equation for the dissolution of NH4NO3 in water. If 4.98 mol of NH4NO3 are dissolved, how many total moles of ions are produced?
each molecule of ammonium nitrate yields one ammonium ion and one nitrate ion
so, twice the number of moles that you started with
The equation for the dissolution of NH4NO3 in water can be written as:
NH4NO3 (s) → NH4+ (aq) + NO3- (aq)
In this equation, NH4NO3 solid dissociates into NH4+ and NO3- ions when dissolved in water.
To calculate the total moles of ions produced when 4.98 mol of NH4NO3 are dissolved, we look at the stoichiometry of the equation. From the equation, we can see that 1 mole of NH4NO3 produces 1 mole of NH4+ ions and 1 mole of NO3- ions.
Therefore, when 4.98 mol of NH4NO3 are dissolved, the same number of moles of NH4+ and NO3- ions are produced. Hence, the total moles of ions produced would be:
Number of moles of ions = 4.98 mol NH4+ ions + 4.98 mol NO3- ions = 9.96 mol of ions
The equation for the dissolution of NH4NO3 in water is:
NH4NO3(s) → NH4+(aq) + NO3-(aq)
In this equation, NH4NO3 breaks down into NH4+ ions and NO3- ions when dissolved in water.
To calculate the total moles of ions produced from 4.98 mol of NH4NO3, we need to look at the ratio of ions produced in the equation.
From the balanced equation, we can see that for every one mole of NH4NO3, one mole of NH4+ ions and one mole of NO3- ions are produced.
Therefore, if we have 4.98 mol of NH4NO3, we will have an equal number of moles of NH4+ ions and NO3- ions.
Therefore, the total moles of ions produced from 4.98 mol of NH4NO3 would be:
Total moles of ions = 4.98 mol (NH4+) + 4.98 mol (NO3-)
Total moles of ions = 9.96 mol