consider the following chemical reactionsna2so3s 2hclaq so2g 2naclaq h2ol 1 nahso3 s hclaq so2g

Step 1: Calculate the volume of sulfur dioxide produced from the reaction involving 1.22 g of so

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Question

Consider the following chemical reactions: Na2SO3(s) + 2HCl(aq) â†’ SO2(g) + 2NaCl(aq) + H2O(l) (1) NaHSO3(s) + HCl(aq) â†’ SO2(g) + NaCl(aq) + H2O(l) (2) 1.22 g of sodium sulfite reacts with excess hydrochloric acid. 0.510 g of sodium hydrogen sulfite reacts with hydrochloric acid as well. Calculate the total volume of sulfur dioxide from reactions (1) and (2) at STP. The molar masses are: Na = 22.99 g/mol, S = 32.00 g/mol, O = 16.00 g/mol, H = 1.008 g/mol.

          
            Consider the following chemical reactions:
Na2SO3(s) + 2HCl(aq) â†’ SO2(g) + 2NaCl(aq) + H2O(l)    (1)
NaHSO3(s) + HCl(aq) â†’ SO2(g) + NaCl(aq) + H2O(l)    (2)

1.22 g of sodium sulfite reacts with excess hydrochloric acid. 0.510 g of sodium hydrogen sulfite reacts with hydrochloric acid as well. Calculate the total volume of sulfur dioxide from reactions (1) and (2) at STP. The molar masses are: Na = 22.99 g/mol, S = 32.00 g/mol, O = 16.00 g/mol, H = 1.008 g/mol.

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Step 1

22 g of sodium sulfite. Given: - Mass of sodium sulfite = 1.22 g - Molar mass of Na2SO3 = 125.98 g/mol - Molar ratio of Na2SO3 to SO2 = 1:1 - 1 mole of any gas at STP = 22.4 L Calculate the volume of sulfur dioxide: \[ \text{Moles of Na2SO3} = \frac{1.22 \text{ Show more…

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Transcript

00:01 So for this problem, we have two reactions occurring.

00:03 We only have one reactant because it's happening with excess hydrochloric acid.

00:07 We're supposed to determine the amount of salt, or hydrogen sulf.

00:11 Nope, not hydrogen sulfide.

00:12 Excuse me.

00:13 We're supposed to figure out the amount of sulfur dioxide.

00:16 So to do that, we're going to do three -step conversion for each of these, converting from grams of the reactant to liters of the product for each of them, and then we'll add them up at the end.

00:28 So the first one, if we start with 1 .22 grams of sodium sulfite.

00:38 The first thing we're going to do is get rid of those grams of sodium sulfate by putting grams of sodium sulfate in the bottom.

00:44 I just realized i'm totally going to run out of room.

00:47 So we're going to start that again.

00:49 So if we start with 1 .22 grams of sodium sulfite, we're going to put grams of sodium sulfite on the bottom.

01:01 And we're going to put one mole of sodium sulfite out top.

01:08 So the molar mass of sodium sulfate i found earlier using periodic table is 125 .98.

01:16 The second step then is to use our molar ratios.

01:20 So we're going to go from moles of na2 -s -o -3 to moles of so2.

01:29 When we do mole on top of mole, we need to look at the coefficients.

01:32 The coefficient of so2 is one, while the coefficient of na2 -s -o -3 is also one.

01:38 Last but not least, we know that one mole of any gas at stp equals 22 .4 liters of that gas at stp.

01:50 So now we're going to multiply through...