This problem is providing information about the initial mass of mercury (II) oxide (10.00 g) which is able to produce liquid mercury (8.00 g) and gaseous oxygen and asks for the resulting mass of the latter, which turns out to be 0.65 g after doing the corresponding calculations.

Initially, it is given a mass of 10.00 g of the oxide and 1.35 g are left which means that the following mass is consumed:

$m_{HgO}^{consumed}=10.00g-1.35 g=8.65 g$

Now, since 8.00 grams of liquid mercury are collected, it is possible to calculate the grams of oxygen that were produced, by considering the law of conservation of mass, which states that the mass of the products equal that of the reactants as it is nor destroyed nor created. In such a way, the mass of oxygen turns out to be:

$m_{O_2}=8.65g-8.00g=0.65g$

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2 years ago

3NO2 + H2O = 2 HNO3 + NO

ArbitrLikvidat [17]

Explanation:

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