Question

6. Given the reaction represented by the balanced equation: Sn(s) + 2 HF (aq) - SnF2(s) + H2(8) a. Calculate the number of grams of SnF2 that can be produced theoretically by mixing 10.00 g of Sn with excess HF. b. If only 11.00 g of SnF, were produced, calculate the percent yield,

Answer 1

a.

Number of moles = Mass in gram / Atomic weight

Atomic weight of Sn = 118.71 g/mol

Number of moles of Sn = 10.00 g / 118.71 g/mol = 0.0842 mol

From reaction, 1.0 mol of Sn produces 1.0 mol of SnF2 then 0.0842 mol of Sn will produce 0.0842 mol of SnF2.

Molar mass of SnF2 = 156.707 g/mol

Mass of SnF2 can be produced = 156.707 g/mol * 0.0842 mol = 13.2 g

b.

Percent yield = Actual yield * 100 / Theoretical yield

Percent yield = 11.00 g * 100 / 13.2 g = 83.33%

Percent yield = 83.33%