# If 1.50 g of H2C2O4.2H2O were heated to drive off the water of hydration, how much anhydrous H2C2O4 would remain?a. 0.34gb. 0.92gc. 1.07gd…. If 1.50 g of H2C2O4.2H2O were heated to drive off the water of hydration, how much anhydrous H2C2O4 would remain?a. 0.34gb. 0.92gc. 1.07gd….

If 1.50 g of H2C2O4.2H2O were heated to drive off the water of hydration, how much anhydrous H2C2O4 would remain?a. 0.34gb. 0.92gc. 1.07gd….

If 1.50 g of H2C2O4.2H2O were heated to drive off the water of hydration, how much anhydrous H2C2O4 would remain?a. 0.34gb. 0.92gc. 1.07gd….
Lets calculate the molecular weight of the hydrated chemical, H2C2O4.2H2O.using the molecular masses of O=16, C=12 and H=1 gm/molar, we getMolecular mass of the hydrated chemical = 2×1+2×12+4×16 + 2x2x1+2×16 = 126 gm/moleand the molecular mass of water of hydration = 2x2x1 + 2×16 = 36 gm/molenow for every 126 g hydrated chemical heated, 36 gm of water would be lost.or, for 1.5 gm hydrated chemical, lost water = 36×1.5/126 gm = 0.429 gmThus, the amount of anhydrous chemical (H2C2O4) left = 1.5-0.429 gm = 1.071 gm Therefore, the correct answer is option C: 1.071 gm.