Determine the quantity (g) of pure CaCl2 in 7.5 g of CaCl2•9H2O.
7.5g CaCl2 * 9H2O * 1molecacl2 * 9h2o/ 273.12 grams = .027moles 0.27moles CaCl2 *9H20 * 110.98g cacl2/ 1 mole CaCl2= 2.9964 grams, with correct Sig Fig, 2.9 Grams wondering if someone can help double check my answer here, a little unsure of my calculations.