Item 1 Constarts 1 Periodi Tatle Cakculating the volume of preduct produced from

Question

Item 1 Constarts 1 Periodi Tatle Cakculating the volume of preduct produced from a given mass of react can be achieved in three sleps 1 Convert the mass of reactant to moles of reactant 2 Use the bolanced chemical equation to determine the number of moles of product prodaced 3 Convent he moles of preduct produced to volumse ing the convension fachor 1 maol ideal gas 22.4 L ideal gae when the gas is at standar temperature and pressare (STP) You will need three cenversion factors to complete the cakculation the molar mass of the reactant the male ratio between reactant and product and the molar vokme of the gas -Part A In an autemobile collsion, sedum acide, NaN, decomponses and 1h an air bag with niogen as equation for the reaction is Express your answer with the appropriate units View Available Hints) Submit

Explanation / Answer

A) Note that the molar mass of NaN3 is 65.01 g/mol. Thus, 125 g of NaN3 corresponds to 125/65.01 = 1.92 mol NaN3.

From the balanced equation, 1 mole of NaN3 produces 1.5 moles of N2(g). So,

(1.5)(1.92) = 2.88 mol of N2 forms.

Since 1 mol of any gas at STP occupies 22.4 L. So, Volume of N2 produced is

2.88*22.4 = 64.61 L of N2 are produced.

B) The balance reaction is: 2Na + 2H2O ---> H2 + 2NaOHFrom the equation: 2mol of Na gives 1 mol H2
Note that the atomic mass of Na is 23.00 g/mol. Thus, 1.60 g of Na corresponds to 1.60/23 = 0.07 mol Na.

From the balanced equation, (1/2)(0.07) = 0.035 mol of H2 forms.

Since 1 mol of any gas at STP occupies 22.41 L.

So volume of H2 produced is 0.035*22.4 = 0.78 L of H2 are produced.

I hope this helps!

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