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What mass of sodium nitrate is required to give #128*g# of oxygen gas upon thermolysis of the salt?

Answer 1

Naomi Aronson

Approx. #700*g#..........

A stoichiometric equation is required:

#NaNO_3(s) + Delta rarr NaNO_2(s) + 1/2O_2(g)#

We require a #128*g# mass of dioxygen gas, i.e. #(128*g)/(32.00*g*mol^-1)=4*mol#.

Given the equation, clearly we need AT LEAST #8*mol# of #"sodium nitrate"#, i.e. #8*molxx85.00*g*mol^-1=??g#

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Answer 2

Cooper Anderson

The molar mass of sodium nitrate (NaNO3) is 85 grams per mole. The molar mass of oxygen (O2) is 32 grams per mole. To calculate the mass of sodium nitrate required to produce 128 grams of oxygen gas, we need to use stoichiometry.

1 mole of NaNO3 produces 3 moles of O2.

Mass of O2 = 128 g Molar mass of O2 = 32 g/mol

Using the stoichiometry:

(128 g O2) * (1 mol NaNO3 / 3 mol O2) * (85 g NaNO3 / 1 mol NaNO3) = 362.67 g NaNO3

So, 362.67 grams of sodium nitrate are required to produce 128 grams of oxygen gas upon thermolysis.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.