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allochka39001 [22]
3 years ago
6

An antacid tablet weighing 0.853g contained calcium carbonate as the active ingredient, in addition to an inert binder. When an

acid solution weighing 56.519g was added to the tablet, carbon dioxide gas was released, producing a fizz. The resulting solution weighed 57.152g. How many grams of carbon dioxide were produced?
Chemistry
1 answer:
ArbitrLikvidat [17]3 years ago
5 0

Answer:

0.22 g of CO2 were produced.

Explanation:

First, let's represent what is happening with an hypothetical chemical equation just to have a clearer vision of the presented process:

CaCO3 (aq) + 2 HAc (aq) → CaAc2 (aq) + H2O (l] + CO2 (g)

We have a tablet that has CaCO3 as the active ingredient that when combined with an acid, in this case represented as HAc, reacts producing a Calcium salt, water and carbon dioxide that will leave the solution as gas.

Having said that, we know that the initial mass of the reactants will have to maintain during the chemical reaction, or what is the same, the quantity of matter during the process will not change. So, if we have a tablet that weighs 0.853 g and we add an acid solution of 56.519 g, then we have that the initial mass of the reactants will be:

0.853 g from tablet + 56.519 g from acid solution = 57.372 g

This amount of matter should be the same after the reaction, but we know that the CO2 gas will leave the solution once it's formed, so considering that the resulting solution weighs 57.152 g we could calculate the mass of CO2 produced:

57.372 g of initial mass - 57.152 g of resulting solution = 0.22 g of CO2 that left the aqueous solution as gas.

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Step1. You must first determine the number of moles that are in 4000 g of octane, using the molar mass of octane. Step 2. Then you must determine the number of moles of carbon dioxide that can be produced by that number of moles of octane, based on the mole ratio between octane and carbon dioxide in the balanced equation. Step 3. Then use the ideal gas law to determine the volume in liters of carbon dioxide that can be formed.

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Using the following equation 5KNO2+2KMnO4+3H2SO4=5KNO3+2MnSO4+K2SO4+3H20. Starting with 2.47 grams of KNO2 and excess KMnO4 how
Aleonysh [2.5K]

Given equation : 5KNO_{2} + 2KMnO_{4} + 3H_{2}SO_{4}\rightarrow 5KNO_{3} + 2MnSO_{4} + K_{2}SO_{4} + 3H_{2}O

Given information = 2.47 grams KNO2 and excess KMnO4 and we need to find grams of water (H2O).

Since KMnO4 is in excess, so grams of water(H2O) can be calculated using grams of KNO2 with the help of stoichiometry.

To find grams of water(H2O) from grams of KNO2 , we need to follow three steps.

Step 1. Convert 2.47 grams of KNO2 to moles of KNO2.

Moles = \frac{grams}{molar mass}

Molar mass of KNO2 = 85.10 g/mol

Moles = 2.47 gram KNO2\times \frac{1 mol KNO2}{85.10 gram KNO2}

Moles = 0.0290 mol KNO2

Step 2. Convert moles of KNO2 to moles of H2O using mole ratio.

Mole ratio are the coefficient present in front of the compound in the balanced equation.

Mole ratio of KNO2 : H2O is 5 : 3 (5 coefficient of KNO2 and 3 coefficient of H2O)

0.0290 mol KNO2\times \frac{3 mol H2O}{5 mol KNO2}

Mole = 0.0174 mol H2O

Step 3. Convert mole of H2O to grams of H2O

Grams = Moles X molar mass

Molar mass of H2O = 18.00 g/mol

Grams = 0.0174 mol H2O\times \frac{18 g H2O}{1 mol H2O}

Grams of water = 0.313 grams H2O

Summary : The above three steps can also be done in a singe setup as shown below.

2.47 gram KNO2\times \frac{(1 mol KNO2)}{(85.10 gram KNO2)}\times \frac{(3 mol H2O)}{(5 mol KNO2)}\times \frac{(18.00 gram H2O)}{(1mol H2O)}

In the above setup similar units get cancelled out and we will get grams of H2O as 0.313 grams water (H2O)

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