There are 4 moles of spectator ions that remain in solution.
The equation of the reaction is;
Na2CO3(aq) + Pb(NO3)2(aq) -------> PbCO3(s) + 2NaNO3(aq)
We have to determine the limiting reactant. This is the reactant that yields the least amount of product. Note that the spectator ions are Na^+ and NO3^- that form NaNO3.
For Na2CO3
1 mole of Na2CO3 yields 2 moles of NaNO3
3 moles of Na2CO3 yields 3 × 2/1 = 6 moles of NaNO3
For Pb(NO3)2
1 mole of Pb(NO3)2 yields 2 moles of NaNO3
2 moles of Pb(NO3)2 yields 2 × 2/1 = 4 moles of NaNO3
We can see that Pb(NO3)2 is the limiting reactant.
Since [NaNO3] = [Na^+] = [NO3^-], it follows that there are 4 moles of spectator ions that remain in solution.
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Answer:
Explanation: Q1 = mc(ice) ΔT (ice warms)
Q2 = ms (ice melts)
Q3 = mc((water) ΔT (water warms)
Q4 = mr (water boils)
Q5 = mc(vapour)ΔT
Answer:
66.2 % of O
Explanation:
Our compound is the lithium nitrite.
LiNO₂
This salt is ionic and can be dissociated: LiNO₂ → Li⁺ + NO₂⁻
We determine the molar mass:
molar mass of Li + 3 . molar mass of N + 6 . molar mass of O
6.94 g/mol + 3. 14 g/mol + 6 . 16 g/mol = 144.94 g/mol
The mass of oxygen contained in 1 mol of lithium nitrite is:
6 . 16 g/mol = 96 g
So the percentage of oxygen present is:
(96 g / 144.94 g) . 100 = 66.2 %
