Question

10 mL of hydrogen combine with 5 mL of oxygen to yield water. When 200 mL of hydrogen at NTP is passed over heated CuO, the CuO loses 0.144 g of its mass. Do these data correspond to the law of constant proportions?

Answer 1

1st Case:
Mass of 10 mL hydrogen at NTP $=\frac{2}{22400}\times 10=0.00089g$

Mass of 5 mL of oxygen at NTP $=\frac{32}{22400}\times 5=0.00714g$

Ratio of masses of hydrogen and oxygen $=0.00089:0.00714=1:8$
2nd Case:
Mass of 200 mt. of hydrogen at NTP = $=\frac{2}{22400}\times 200=0.0178g$

Mass of oxygen lost by CuO which actually combined with 0.0178 g hydrogen at NTP to form water = 0.144 g Ratio of masses of hydrogen and oxygen = 0.0178 : 0.144 = 1: 8 Hence, law of constant proportions is proved.