What volume of dry silicon tetrafluoride (at STP) can be produced by the complete reaction of 89g og CaSiO?

To determine the volume of dry silicon tetrafluoride (SiF₄) produced from the complete reaction of 89 g of calcium silicate (CaSiO₃), we first need to consider the balanced chemical reaction:

[ \text{CaSiO}_3 + 4\text{HF} \rightarrow \text{SiF}_4 + \text{CaF}_2 + 3\text{H}_2\text{O} ]

Calculating the molar mass of CaSiO₃ (approximately 116.16 g/mol), we find that 89 g corresponds to about 0.766 moles of CaSiO₃. According to the reaction stoichiometry, 1 mole of CaSiO₃ produces 1 mole of SiF₄, yielding around 0.766 moles of SiF₄. At STP, 1 mole of any gas occupies 22.4 liters, so the volume of SiF₄ produced is approximately ( 0.766 , \text{moles} \times 22.4 , \text{L/mole} \approx 17.2 , \text{L} ).