calculate the mass of lead || trioxonitrate(v) that will react with 7.8g of sodium chloride to form 10.5g of sodium trioxonitrate (v). if the law of conservation of mass is obeyed
That chemical with the funny name is properly called lead(II) nitrate.
Pb(NO3)2 + 2NaCl ==> PbCl2 + 2NaNO3.
mols PbCl2 needed = grams/molar mass = ?
You get 1 mol PbCl2 for every 1 mol Pb(NO3)2 used initially.
Then grams Pb(NO3)2 = mols Pb(NO3)2 x molar mass Pb(NO3)2 = ?
I like the answer by anonymous, especially about that funny name; however, anonymous didn't answer the question as the question asks. That is the questions asks to be answered using the law of conservation of mass. Note, however, that there isn't enough information given to answer the question using the law of conservation of mass. It would be done this way.
............Pb(NO3)2 + 2NaCl ==> PbCl2 + 2NaNO3.
................x grams.......7.8 g..........?............10.5 g.
g Pb(NO3)2 + g NaCl = g NaCl + g NaNO3 and solve for x.
BUT, since there is no mass given for PbCl2, the problem can't be answered this way.
To calculate the mass of lead(II) trioxonitrate(V), we need to set up a balanced chemical equation for the reaction and use stoichiometry.
The balanced chemical equation for the reaction can be written as:
2Pb(NO3)2 + 6NaCl → 2NaNO3 + PbCl2
From the balanced equation, we can see that 2 moles of Pb(NO3)2 react with 6 moles of NaCl to form 2 moles of NaNO3 and 1 mole of PbCl2.
First, let's calculate the number of moles of NaCl and NaNO3:
Molar mass of NaCl = 22.99 g/mol + 35.45 g/mol = 58.44 g/mol
Number of moles of NaCl = mass of NaCl / molar mass of NaCl
Number of moles of NaCl = 7.8 g / 58.44 g/mol ≈ 0.1334 mol
Molar mass of NaNO3 = 22.99 g/mol + 14.01 g/mol + (3 * 16.00 g/mol) = 85.00 g/mol
Number of moles of NaNO3 = mass of NaNO3 / molar mass of NaNO3
Number of moles of NaNO3 = 10.5 g / 85.00 g/mol ≈ 0.1235 mol
Since the reaction ratio shows that 2 moles of Pb(NO3)2 react with 2 moles of NaNO3, we can say that 0.1235 mol of NaNO3 corresponds to 0.1235 mol of Pb(NO3)2.
Now, let's calculate the mass of Pb(NO3)2:
Molar mass of Pb(NO3)2 = 207.2 g/mol + (2 * (14.01 g/mol + (3 * 16.00 g/mol))) = 331.22 g/mol
Mass of Pb(NO3)2 = number of moles of Pb(NO3)2 * molar mass of Pb(NO3)2
Mass of Pb(NO3)2 = 0.1235 mol * 331.22 g/mol ≈ 40.99 g
Therefore, the mass of lead(II) trioxonitrate(V) that will react with 7.8g of sodium chloride to form 10.5g of sodium trioxonitrate(V) is approximately 40.99 grams.
To calculate the mass of lead(II) trioxonitrate(V) (Pb(NO3)2) that will react with the given amount of sodium chloride (NaCl) to form the given amount of sodium trioxonitrate(V) (NaNO3), we need to determine the stoichiometry of the reaction.
First, let's write the balanced chemical equation for the reaction between lead(II) trioxonitrate(V) and sodium chloride:
2Pb(NO3)2 + 2NaCl → 2NaNO3 + PbCl2
From the balanced equation, we can see that 2 moles of lead(II) trioxonitrate(V) react with 2 moles of sodium chloride to produce 2 moles of sodium trioxonitrate(V) and 1 mole of lead(II) chloride.
Next, we calculate the molar mass of sodium chloride (NaCl), which is:
Na = 22.99 g/mol
Cl = 35.45 g/mol
Molar mass of NaCl = (22.99 g/mol) + (35.45 g/mol) = 58.44 g/mol
Now, let's calculate the number of moles of NaCl using the given mass:
moles of NaCl = mass of NaCl / molar mass of NaCl
moles of NaCl = 7.8 g / 58.44 g/mol ≈ 0.1336 mol
According to the stoichiometry of the balanced equation, we can see that each mole of Pb(NO3)2 reacts with 1 mole of NaCl. Therefore, the number of moles of Pb(NO3)2 is also 0.1336 mol.
Finally, we calculate the mass of Pb(NO3)2 using its molar mass:
Molar mass of Pb(NO3)2 = (207.2 g/mol) + [(14.01 g/mol) + (3 * 16.00 g/mol)] * 2
Molar mass of Pb(NO3)2 ≈ 331.2 g/mol
mass of Pb(NO3)2 = moles of Pb(NO3)2 * molar mass of Pb(NO3)2
mass of Pb(NO3)2 = 0.1336 mol * 331.2 g/mol ≈ 44.2 g
Therefore, the mass of lead(II) trioxonitrate(V) that will react with 7.8 g of sodium chloride to form 10.5 g of sodium trioxonitrate(V) is approximately 44.2 grams.