Final answer:
A toothpaste manufacturer can expect to get 234.85 grams of SnF2 when reacting 30.00 grams of HF with enough tin, determined through stoichiometric calculations based on the balanced chemical equation.
Step-by-step explanation:
To determine how many grams of SnF2 can be produced from 30.00 grams of HF, we need to perform a series of stoichiometric calculations based on the balanced chemical equation Sn(s) + 2HF(g) → SnF2(s) + H2(g). First, we find the molar mass of HF (20.01 g/mol) and then convert grams to moles to find the moles of HF. Next, using the stoichiometric ratio of 1:1 between SnF2 and HF from the balanced equation, we find the moles of SnF2 that can be formed. Finally, we convert the moles of SnF2 back to grams using its molar mass (156.69 g/mol).
Step 1: Calculate moles of HF:
(30.00 g HF) / (20.01 g/mol HF) = 1.499 moles of HF
Step 2: Convert moles of HF to moles of SnF2:
Because the ratio is 1:1, 1.499 moles of HF will produce 1.499 moles of SnF2.
Step 3: Convert moles of SnF2 to grams of SnF2:
(1.499 moles SnF2) × (156.69 g/mol SnF2) = 234.85 grams of SnF2
Therefore, the toothpaste manufacturer can expect to get 234.85 grams of SnF2.