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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The proton transfer reaction between Cyanide and water can be written as; X^- + H2O -----> HX + OH^-
<h3>What is a proton transfer reaction?</h3>
A proton transfer reaction is one in which a proton is moved from one chemical specie to another.It is in fact and acid - base reaction in the Brownstead - Lowry sense.
The proton transfer reaction between Cyanide and water can be written as(Let the cyanide ion be shown as X);
X^- + H2O -----> HX + OH^-
Learn more about proton transfer: brainly.com/question/861100?
The reaction for the combustion of methane can be expressed as follows.
CH4 + 2O2 --> CO2 + 2H2O
We solve first for the amount of carbon dioxide in moles by dividing the given volume by 22.4L which is the volume of 1 mole of gas at STP.
moles of CO2 = (5.6 L) / (22.4 L/1 mole)
moles of CO2 = 0.25 moles
Then, we can see that every mole of carbon dioxide will need 1 mole of methane
moles methane = (0.25 moles CO2) x (1 moles O2/1 mole CO2)
= 0.25 moles CH4
Then, multiply this by the molar mass of methane which is 16 g/mole. Thus, the answer is 4 grams methane.
