Can you check my answer?When sodium chloride reacts with silver nitrate, silver chloride precipitates. What mass of AgCl is produced from 75.0 g AgNO3?
My answer: 63.3g AgCl
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To check your answer, we need to use stoichiometry and the molar masses of the substances involved.
1. Write the balanced chemical equation for the reaction:
NaCl + AgNO3 -> AgCl + NaNO3
2. Calculate the molar mass of AgNO3:
AgNO3 = 107.87 g/mol + 14.01 g/mol + (3 * 16.00 g/mol) = 169.87 g/mol
3. Convert the given mass of AgNO3 into moles:
75.0 g AgNO3 * (1 mol AgNO3 / 169.87 g AgNO3) = 0.4416 mol AgNO3
4. Use stoichiometry to determine the number of moles of AgCl produced:
Using the balanced equation, the stoichiometric ratio between AgNO3 and AgCl is 1:1.
So, the moles of AgCl produced will be the same as the moles of AgNO3 used: 0.4416 mol
5. Calculate the molar mass of AgCl:
AgCl = 107.87 g/mol + 35.45 g/mol = 143.32 g/mol
6. Convert the moles of AgCl into mass:
0.4416 mol AgCl * (143.32 g AgCl / 1 mol AgCl) = 63.2 g AgCl
Upon checking your answer, it appears that you made a rounding error as you wrote 63.3 g AgCl. The correct mass of AgCl produced from 75.0 g AgNO3 is 63.2 g.