Final answer:
10.8 grams of magnesium react to produce approximately 52.73 grams of tin, based on stoichiometric calculations involving the balanced chemical equation and molar masses.
Step-by-step explanation:
The student has asked how many grams of tin (Sn) is formed when 10.8 grams of magnesium (Mg) reacts with an excess of tin(IV) nitrite, Sn(NO2)4. To solve this problem, we need to conduct a stoichiometric calculation involving the balanced chemical equation:
Mg + Sn(NO2)4 → Sn + Mg(NO2)2.
Assuming magnesium is the limiting reactant, let's calculate the number of moles of Mg (molar mass = 24.305 g/mol) present in 10.8 grams:
moles of Mg = mass of Mg / molar mass of Mg = 10.8 g / 24.305 g/mol ≈ 0.444 moles of Mg.
From the equation, the stoichiometry shows that 1 mole of Mg reacts to produce 1 mole of Sn. Therefore, the number of moles of Sn produced will also be 0.444 mole. Using the molar mass of Sn (118.71 g/mol), we can calculate the mass of Sn formed:
mass of Sn = moles of Sn × molar mass of Sn = 0.444 moles × 118.71 g/mol ≈ 52.73 grams.
Thus, when 10.8 grams of Mg react with an excess of Sn(NO2)4, approximately 52.73 grams of Sn is formed, after rounding to two decimal places.