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What is the volume of oxygen gas at STP from the decomposition of 1.70 g of sodium nitrate (85.00 g/mol)?

Answer :

Final answer:

The decomposition of 1.70 g of sodium nitrate at STP would produce 0.224 liters of oxygen gas.

Explanation:

To determine the volume of oxygen gas at STP from the decomposition of sodium nitrate, we'll first need to convert the given mass of sodium nitrate into moles. We can do this using the molar mass.

We know that one mole of sodium nitrate (NaNO3) weighs 85.00 g, and that 1.70 g of sodium nitrate is therefore (1.70 g / 85.00 g/mol) = 0.020 mol.

Decomposition of sodium nitrate produces oxygen, according to this balanced chemical equation: 2NaNO3(s) → 2NaNO2(s)+O2(g)

From the balanced equation, we can see that 2 moles of NaNO3 produce 1 mole of O2. So 0.020 mol of NaNO3 would produce 0.020/2 = 0.010 mol of O2.

At Standard Temperature and Pressure (STP), one mole of any gas occupies 22.4 liters. Therefore, 0.010 mol of O2 would occupy 0.010 mol x 22.4 L/mol = 0.224 L.

So, the volume of oxygen gas produced by the decomposition of 1.70 g of sodium nitrate at STP is 0.224 liters.

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