Question

When 15. 3 g NaCl reacts with 81.70 g Pb(NO3)2, sodium chloride is the limiting reactant, according to the balanced equation below. How many grams of lead (II) nitrate will remain after the reaction stops?

2 NaCl + 1 Pb(NO3)2 → 2 NaNO3 + 1 PbCl2

Answer 1

Answer:

$m_{Pb(NO_3)_2}^{leftover}=38.34g$

Explanation:

Hello there!

In this case, given the chemical reaction by which the sodium chloride reacts with lead (II) nitrate and the former is the limiting reactant, it is possible to calculate the mass of lead (II) nitrate that are actually consumed according to the 2:1 mole ratio between them:

$m_{Pb(NO_3)_2}=15.3gNaCl*\frac{1molNaCl}{58.44gNaCl}*\frac{1molPb(NO_3)_2}{2molNaCl} *\frac{331.21gPb(NO_3)_2}{1molPb(NO_3)_2} \\\\m_{Pb(NO_3)_2}=43.36gPb(NO_3)_2$

Thus, the leftover of lead (II) nitrate is:

$m_{Pb(NO_3)_2}^{leftover}=81.70g-43.36g\\\\m_{Pb(NO_3)_2}^{leftover}=38.34g$

Best regards!