Answer :
Answer:
[tex] \frac{(179.6g) \times (1mol)}{63g} = 2.8 \approx3 \: mol \: HNO_3\\ \frac{4 mol H_2O \times 3}{8 mol HNO_3}=1.5 mol\\ H_2O\\1.5 mol H2O × 175.6g H_2O / 1 mol H_2O = 263. 4g H_2O [/tex]
Answer:
[tex] \frac{(179.6g) \times (1mol)}{63g} = 2.8 \approx3 \: mol \: HNO_3\\ \frac{4 mol H_2O \times 3}{8 mol HNO_3}=1.5 mol\\ H_2O\\1.5 mol H2O × 175.6g H_2O / 1 mol H_2O = 263. 4g H_2O [/tex]