Heating Soda Mix: Calculate Salt Mass & Gas Volume

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Hey guys! Today, we're diving into a fun chemistry problem involving a mixture of crystalline and baking soda. When we heat this mix, some interesting stuff happens – the mass decreases, and gas is released. Our mission? To figure out the mass of each salt in the original mixture. Let's break it down step by step!

Understanding the Problem

So, we have a mixture of two types of soda: crystalline soda (also known as washing soda, which is hydrated sodium carbonate, (Na_2CO_3 \cdot nH_2O)) and baking soda (sodium bicarbonate, NaHCO3{NaHCO_3}). When we heat this mixture, the baking soda decomposes, releasing carbon dioxide gas and water vapor. The crystalline soda might also lose its water of hydration depending on the temperature. We are given that the mass of the mixture decreases to 31.8 g and 2.24 L of gas (at standard temperature and pressure, or STP) is released. Our goal is to determine the initial masses of both the crystalline and baking soda.

Key Information

  • Initial Mixture: Crystalline soda ((Na_2CO_3 \cdot nH_2O)) and baking soda (NaHCO3{NaHCO_3})
  • Final Mass: 31.8 g
  • Gas Released: 2.24 L (at STP)

Reactions Involved

The main reaction we need to consider is the decomposition of baking soda:

2NaHCO3(s)→Na2CO3(s)+H2O(g)+CO2(g){ 2NaHCO_3(s) \rightarrow Na_2CO_3(s) + H_2O(g) + CO_2(g) }

The crystalline soda may lose water upon heating, depending on the value of n in (Na_2CO_3 \cdot nH_2O). For example, if it's washing soda (Na_2CO_3 \cdot 10H_2O):

Na2CO3⋅10H2O(s)→Na2CO3(s)+10H2O(g){ Na_2CO_3 \cdot 10H_2O(s) \rightarrow Na_2CO_3(s) + 10H_2O(g) }

Step-by-Step Solution

1. Calculate Moles of Gas Released

At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 L. We can use this to find the number of moles of gas released:

Moles of gas=Volume of gasMolar volume at STP=2.24 L22.4 L/mol=0.1 mol{ \text{Moles of gas} = \frac{\text{Volume of gas}}{\text{Molar volume at STP}} = \frac{2.24 \text{ L}}{22.4 \text{ L/mol}} = 0.1 \text{ mol} }

2. Determine Moles of NaHCO3{NaHCO_3} Decomposed

From the balanced equation for the decomposition of NaHCO3{NaHCO_3}, 2 moles of NaHCO3{NaHCO_3} produce 1 mole of CO2{CO_2} and 1 mole of H2O{H_2O}. Therefore, the total moles of gas produced from NaHCO3{NaHCO_3} decomposition is 0.1 mol. Since 1 mole of CO2{CO_2} is produced for every 2 moles of NaHCO3{NaHCO_3} that decompose:

Moles of NaHCO3=2×Moles of gas=2×0.1 mol=0.2 mol{ \text{Moles of } NaHCO_3 = 2 \times \text{Moles of gas} = 2 \times 0.1 \text{ mol} = 0.2 \text{ mol} }

3. Calculate Initial Mass of NaHCO3{NaHCO_3}

The molar mass of NaHCO3{NaHCO_3} (sodium bicarbonate) is:

Molar Mass(NaHCO3)=22.99(Na)+1.01(H)+12.01(C)+3×16.00(O)=84.01 g/mol{ Molar\ Mass(NaHCO_3) = 22.99 (Na) + 1.01 (H) + 12.01 (C) + 3 \times 16.00 (O) = 84.01 \text{ g/mol} }

Now, we can find the initial mass of NaHCO3{NaHCO_3}:

Mass of NaHCO3=Moles of NaHCO3×Molar Mass(NaHCO3)=0.2 mol×84.01 g/mol=16.802 g{ \text{Mass of } NaHCO_3 = \text{Moles of } NaHCO_3 \times Molar\ Mass(NaHCO_3) = 0.2 \text{ mol} \times 84.01 \text{ g/mol} = 16.802 \text{ g} }

4. Determine the Mass Loss from NaHCO3{NaHCO_3} Decomposition

When NaHCO3{NaHCO_3} decomposes, it forms Na2CO3{Na_2CO_3}, H2O{H_2O}, and CO2{CO_2}. The mass loss is due to the release of H2O{H_2O} and CO2{CO_2}. From the reaction equation, 2 moles of NaHCO3{NaHCO_3} decompose to produce 1 mole of H2O{H_2O} and 1 mole of CO2{CO_2}. Thus, 0.2 moles of NaHCO3{NaHCO_3} will produce 0.1 moles each of H2O{H_2O} and CO2{CO_2}.

Let's calculate the masses of H2O{H_2O} and CO2{CO_2}:

  • Molar mass of H2O=2×1.01+16.00=18.02 g/mol{H_2O = 2 \times 1.01 + 16.00 = 18.02 \text{ g/mol}}
  • Mass of H2O=0.1 mol×18.02 g/mol=1.802 g{H_2O = 0.1 \text{ mol} \times 18.02 \text{ g/mol} = 1.802 \text{ g}}
  • Molar mass of CO2=12.01+2×16.00=44.01 g/mol{CO_2 = 12.01 + 2 \times 16.00 = 44.01 \text{ g/mol}}
  • Mass of CO2=0.1 mol×44.01 g/mol=4.401 g{CO_2 = 0.1 \text{ mol} \times 44.01 \text{ g/mol} = 4.401 \text{ g}}

Total mass loss from NaHCO3{NaHCO_3} decomposition:

Mass loss=Mass of H2O+Mass of CO2=1.802 g+4.401 g=6.203 g{ \text{Mass loss} = \text{Mass of } H_2O + \text{Mass of } CO_2 = 1.802 \text{ g} + 4.401 \text{ g} = 6.203 \text{ g} }

5. Calculate the Initial Mass of the Mixture

We know the final mass of the mixture after heating is 31.8 g, and the mass loss is 6.203 g. Therefore, the initial mass of the mixture is:

Initial mass of mixture=Final mass+Mass loss=31.8 g+6.203 g=38.003 g{ \text{Initial mass of mixture} = \text{Final mass} + \text{Mass loss} = 31.8 \text{ g} + 6.203 \text{ g} = 38.003 \text{ g} }

6. Determine the Initial Mass of Crystalline Soda

Now we know the initial mass of the mixture and the initial mass of NaHCO3{NaHCO_3}. We can find the initial mass of the crystalline soda:

Mass of crystalline soda=Initial mass of mixture−Mass of NaHCO3=38.003 g−16.802 g=21.201 g{ \text{Mass of crystalline soda} = \text{Initial mass of mixture} - \text{Mass of } NaHCO_3 = 38.003 \text{ g} - 16.802 \text{ g} = 21.201 \text{ g} }

7. Consider Water Loss from Crystalline Soda

Let's assume the crystalline soda is (Na_2CO_3 \cdot nH_2O). When heated, it loses water. The problem states the mass of the mixture decreased to 31.8 g. This decrease includes both the loss of H2O{H_2O} and CO2{CO_2} from the decomposition of NaHCO3{NaHCO_3} and any water lost from the crystalline soda.

Since we've already accounted for the mass loss from NaHCO3{NaHCO_3}, we need to figure out if the crystalline soda lost water and how much. Let's denote the mass of anhydrous Na2CO3{Na_2CO_3} remaining after heating as mNa2CO3{m_{Na_2CO_3}}.

The final mass of 31.8 g consists of Na2CO3{Na_2CO_3} from both the decomposed NaHCO3{NaHCO_3} and the heated crystalline soda.

First, calculate the mass of Na2CO3{Na_2CO_3} formed from the decomposition of 0.2 mol of NaHCO3{NaHCO_3}:

2NaHCO3(s)→Na2CO3(s)+H2O(g)+CO2(g){ 2NaHCO_3(s) \rightarrow Na_2CO_3(s) + H_2O(g) + CO_2(g) }

So, 0.2 mol of NaHCO3{NaHCO_3} yields 0.1 mol of Na2CO3{Na_2CO_3}.

The molar mass of Na2CO3=2×22.99+12.01+3×16.00=105.99 g/mol{Na_2CO_3 = 2 \times 22.99 + 12.01 + 3 \times 16.00 = 105.99 \text{ g/mol}}.

Thus, the mass of Na2CO3{Na_2CO_3} from NaHCO3{NaHCO_3} is:

Mass of Na2CO3=0.1 mol×105.99 g/mol=10.599 g{ \text{Mass of } Na_2CO_3 = 0.1 \text{ mol} \times 105.99 \text{ g/mol} = 10.599 \text{ g} }

Now, we subtract this from the final mass to find the mass of Na2CO3{Na_2CO_3} that came from the crystalline soda:

mNa2CO3=31.8 g−10.599 g=21.201 g{ m_{Na_2CO_3} = 31.8 \text{ g} - 10.599 \text{ g} = 21.201 \text{ g} }

This is the mass of anhydrous Na2CO3{Na_2CO_3} that remains from the original crystalline soda. Note that this value is exactly the mass we calculated for crystalline soda before considering water loss. This indicates that all the crystalline soda was converted to anhydrous Na2CO3{Na_2CO_3}.

8. Determine the Hydration of the Crystalline Soda

We know the initial mass of the crystalline soda was 21.201 g, and it converted to 21.201 g of anhydrous Na2CO3{Na_2CO_3}. This means the crystalline soda completely dehydrated to Na2CO3{Na_2CO_3}, losing all its water of hydration during the heating process. However, because the mass of the solid Na2CO3{Na_2CO_3} is equal to the mass of the initial crystalline soda, it means that the crystalline soda was already anhydrous Na2CO3{Na_2CO_3} from the start. Thus, n=0.

9. Final Answer

  • Mass of Baking Soda (NaHCO3{NaHCO_3}): 16.802 g
  • Mass of Crystalline Soda (Na2CO3{Na_2CO_3}): 21.201 g

Conclusion

So there you have it! By carefully analyzing the reactions and the mass changes, we determined the initial masses of both the baking soda and the crystalline soda in the mixture. Remember, the key is to break down the problem into smaller, manageable steps and to keep track of what's happening with each component of the mixture. Keep experimenting, and have fun with chemistry!