What is the coefficient of the H2O in the balanced chemical
1 answer:
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Answer:
0.0585 M
Explanation:
- Pb(NO₃)₂ (aq) + 2NaCl (aq) → PbCl₂ (s) + 2NaNO₃ (aq)
First we <u>calculate the inital number of moles of each reagent</u>, using the <em>given volumes and concentrations</em>:
- 0.255 M Pb(NO₃)₂ * 52.1 mL = 13.3 mmol Pb(NO₃)₂
- 0.415 M NaCl * 38.5 mL = 16.0 mmol NaCl
Then we <u>calculate how many Pb(NO₃)₂ moles reacted with 16.0 mmoles of NaCl</u>, using the <em>stoichiometric coefficients of the reaction</em>:
- 16.0 mmol NaCl *
= 8.00 mmol Pb(NO₃)₂
Now we <u>calculate the remaining number of Pb(NO₃)₂ moles after the reaction</u>:
- 13.3 mmol - 8.00 mmol = 5.30 mmol Pb(NO₃)₂
Finally we <em>divide the number of moles by the final volume</em> to <u>calculate the concentration</u>:
- 5.30 mmol / (52.1 mL + 38.5 mL) = 0.0585 M