What volume of gas is generated at STP when 4.6 g of Aluminum Carbonate is placed in 0.75 M Nitric acid?

Question

Wyatt Atkins · Accepted Answer

You'd produce #"1.3 L"# of #CO_2# gas at STP with that much aluminium carbonate.

SIDE NOTE: As you did not specify the volume of the nitric acid, I will assume that it is a 1-L sample. This assumption is based on the knowledge that nitric acid will not function as a limiting reagent, meaning that there will be excess nitric acid.

Thus, begin with the chemical equation that is balanced.

#Al_2(CO_3)_(3(aq)) + 6HNO_3(aq) -> 2Al(NO_3)_(3(aq)) + 3H_2O_((l)) + 3CO_(2(g))#

Notice that you have a #"1:3"# mole ratio between aluminium carbonate and carbon dioxide, i.e. you'll produce three times as many moles of the latter than the number of moles of the former that react.

The quantity of aluminum carbonate in moles is

#"4.6 g" * "1 mole"/"234.0 g" = "0.0197 moles"#

Currently, a 1.0-liter nitric acid sample will have

#C = n/V => n = C * V = "0.75 M" * "1.0 L" = "0.75 moles nitric acid"#

which is more than sufficient to confirm that the reaction is not being limited by the nitric acid.

According to the #"1:6"# mole ratio that exists between aluminium carbonate and nitric acid, you would need a minimum of

#"0.0197 moles aluminijm carbonate" * "6 moles nitric acid"/"1 mole aluminium carbonate" = "0.118 moles nitric acid"#

You do in fact have more nitric acid than you require in this instance.

To calculate how many moles of carbon dioxide will be produced, use their respective mole ratios.

#"0.0197 moles aluminium carbonate" * "3 moles carbon dioxide"/"1 mole aluminium carbonate" = "0.0591 moles"#

The molar volume of a gas at STP is defined as 1 mole of any ideal gas occupying precisely 22.4 L at 273,15 K and 1 atm.

As a result, the amount of gas generated will be

#n_(CO_2) = V/V_("molar") => V = n_(CO_2) * V_("molar")#

#V = "0.0591 moles" * "22.4 L"/"1 mole" = "1.32 L"#

The answer, rounded to two significance figures, will be the same for both 4.6 g and 0.75 M.

#V = "1.3L"#

Note: If a volume for nitric acid is given, find out how many moles of nitric acid you would need in that volume and compare it to the minimum number of moles needed for that amount of aluminum carbonate. If the number is less, the nitric acid will act as a limiting reagent and you will need to recalculate the number of moles of aluminum carbonate that react.

Jose Ayala · Answer

undefined To determine the volume of gas generated at STP when 4.6 g of Aluminum Carbonate is placed in 0.75 M Nitric acid, we need to first calculate the moles of Aluminum Carbonate reacted using its molar mass. Then, we use the stoichiometry of the reaction between Aluminum Carbonate and Nitric acid to find the moles of gas produced. Finally, we apply the ideal gas law to calculate the volume of gas at STP.

The balanced chemical equation for the reaction is:

$$Al2(CO3)3 + 6HNO3 → 2Al(NO3)3 + 3CO2 + 3H2O$$

Now, let's proceed with the calculations:

1. Calculate moles of Aluminum Carbonate ($Al2(CO3)3$):

$$\text{Molar mass of } Al2(CO3)3 = 2(26.98) + 3(12.01 + 3(16.00)) = 2(26.98) + 3(12.01 + 3(16.00)) = 291.03 \, \text{g/mol}$$

$$\text{Moles of } Al2(CO3)3 = \frac{4.6 \, \text{g}}{291.03 \, \text{g/mol}} = 0.0158 \, \text{mol}$$

2. According to the balanced equation, 1 mole of Aluminum Carbonate produces 3 moles of gas ($CO2$):

$$\text{Moles of gas produced} = 0.0158 \, \text{mol} \times 3 = 0.0474 \, \text{mol}$$

3. Now, we apply the ideal gas law to find the volume of gas at STP:

$$PV = nRT$$

Where:
- $P$ = pressure (STP = 1 atm)
- $V$ = volume (what we're solving for)
- $n$ = moles of gas
- $R$ = ideal gas constant ($0.0821 \, \text{atm} \cdot \text{L/mol} \cdot \text{K}$)
- $T$ = temperature (STP = 273 K)

$$V = \frac{nRT}{P} = \frac{(0.0474 \, \text{mol})(0.0821 \, \text{atm} \cdot \text{L/mol} \cdot \text{K})(273 \, \text{K})}{1 \, \text{atm}}$$

$$V \approx 1.18 \, \text{L}$$

Therefore, approximately 1.18 liters of gas is generated at STP when 4.6 g of Aluminum Carbonate is placed in 0.75 M Nitric acid.