AN an antacid tablet contains 0.1g of magnesium hydrocarbonate, what nass of stomach acid, HCL, will it neutralise?

Wine is made by the fermentation of the sugar in grapes.
Surose-> Ethanol+Carbondioxide.

Whta massof ethanol can be obtained from 6.00kg of sucrose?

You would do better to post one problem per post.

#1.Write the equation.
Mg(HCO3)2 + 2HCl==>MgCl2 + 2H2O + 2CO2
Convert 0.1 g Mg(HCO3)2 to mols.
Convert mols Mg(HCO3)2 to mols HCl using the coefficients in the balanced equation.
Convert mols HCl to g using g = mols HCl x molar mass HCl.

#2. Write the equation.
C6H12O6 ==>2C2H5OH + 2CO2
Convert 10 kg sucrose to mols.
Convert mols sucrose to mols ethanol using the coefficients in the balanced equation.
Convert mols ethanol to g ethanol using mass = mols ethanol x molar mass ethanol.
Check my work. Check my thinking.

Given all solution of my question

To determine the mass of stomach acid (HCl) that will be neutralized by the antacid tablet containing 0.1g of magnesium hydrocarbonate, we need to consider the balanced chemical equation for the reaction between magnesium hydrocarbonate and hydrochloric acid.

The balanced chemical equation is:

Mg(HCO3)2 + 2HCl → MgCl2 + 2CO2 + 2H2O

From the equation, we can see that 1 mole of magnesium hydrocarbonate (Mg(HCO3)2) reacts with 2 moles of hydrochloric acid (HCl).

To calculate the mass of HCl that will be neutralized, we need to use stoichiometry.

1. Calculate the molar mass of Mg(HCO3)2:
The molar mass of magnesium hydrocarbonate (Mg(HCO3)2) can be calculated by summing up the atomic masses of its elements:
Mg(HCO3)2 = 1Mg + 2(H) + 2(C) + 6(O)
= 24.305 + 2(1.008) + 2(12.011) + 6(15.999)
= 84.305 g/mol

2. Determine the number of moles of Mg(HCO3)2:
Using the given mass of 0.1g and the calculated molar mass of Mg(HCO3)2:
moles = mass / molar mass
moles = 0.1g / 84.305 g/mol
moles ≈ 0.001186 mol

3. Calculate the number of moles of HCl:
Since the ratio of the balanced chemical equation is 1:2, the number of moles of HCl is twice the number of moles of Mg(HCO3)2:
moles of HCl = 2 * 0.001186 mol
moles of HCl ≈ 0.002372 mol

4. Convert moles of HCl to mass:
To convert moles to mass, we need to use the molar mass of HCl, which is approximately 36.461 g/mol:
mass = moles * molar mass
mass = 0.002372 mol * 36.461 g/mol
mass ≈ 0.0864 g

Therefore, the antacid tablet containing 0.1g of magnesium hydrocarbonate will neutralize approximately 0.0864g of stomach acid (HCl).

Moving on to the second question,
To determine the mass of ethanol that can be obtained from 6.00kg of sucrose, we need to use the molar mass of sucrose and the balanced equation for the fermentation reaction.

The balanced equation for the fermentation of sucrose is:
C12H22O11 → 4C2H5OH + 4CO2

From the equation, we can see that 1 mole of sucrose (C12H22O11) produces 4 moles of ethanol (C2H5OH).

1. Calculate the molar mass of sucrose (C12H22O11):
The molar mass of sucrose can be calculated by summing up the atomic masses of its elements:
C12H22O11 = 12(C) + 22(H) + 11(O)
= 12(12.011) + 22(1.008) + 11(15.999)
= 342.297 g/mol

2. Determine the number of moles of sucrose:
Using the given mass of 6.00kg and the calculated molar mass of sucrose:
moles = mass / molar mass
moles = 6000g / 342.297 g/mol
moles ≈ 17.541 mol

3. Calculate the number of moles of ethanol:
Since the ratio of the balanced chemical equation is 1:4, the number of moles of ethanol is four times the number of moles of sucrose:
moles of ethanol = 4 * 17.541 mol
moles of ethanol ≈ 70.164 mol

4. Convert moles of ethanol to mass:
To convert moles to mass, we need to use the molar mass of ethanol, which is approximately 46.07 g/mol:
mass = moles * molar mass
mass = 70.164 mol * 46.07 g/mol
mass ≈ 3229.68 g

Therefore, from 6.00kg of sucrose, approximately 3229.68g (or 3.23kg) of ethanol can be obtained through fermentation.

To determine the mass of stomach acid (HCl) that can be neutralized by the antacid tablet containing 0.1g of magnesium hydrocarbonate, we need to know the molar mass and the balanced chemical equation for the reaction between magnesium hydrocarbonate and hydrochloric acid.

The balanced chemical equation for the reaction between magnesium hydrocarbonate and hydrochloric acid can be represented as follows:

Mg(HCO3)2 + 2HCl -> 2CO2 + 2H2O + MgCl2

To find the molar mass of magnesium hydrocarbonate, we add up the atomic masses of its constituent elements:
Molar mass of Mg(HCO3)2 = (1 x Mg) + (2 x H) + (2 x C) + (6 x O)
= 24.31 + 2 + (2 x 12.01) + (6 x 16.00)
= 24.31 + 2 + 24.02 + 96.00
= 146.34 g/mol

Next, we need to convert the given mass of magnesium hydrocarbonate (0.1g) to moles using its molar mass:
Moles of magnesium hydrocarbonate = Given mass / Molar mass
= 0.1g / 146.34 g/mol
= 0.0006822 mol

From the balanced chemical equation, we can see that the stoichiometric ratio between magnesium hydrocarbonate and hydrochloric acid is 1:2. This means for each mole of magnesium hydrocarbonate, two moles of hydrochloric acid are required.

Hence, the moles of hydrochloric acid that can be neutralized by the antacid tablet containing 0.1g of magnesium hydrocarbonate is:
Moles of HCl = 2 x Moles of magnesium hydrocarbonate
= 2 x 0.0006822 mol
= 0.0013644 mol

Finally, to find the mass of hydrochloric acid that can be neutralized, we multiply the moles of HCl by its molar mass:
Mass of HCl = Moles of HCl x Molar mass of HCl
= 0.0013644 mol x (1 x 1.01 + 1 x 35.45) g/mol
= 0.0013644 mol x 36.46 g/mol
= 0.0498 g

Therefore, the antacid tablet containing 0.1g of magnesium hydrocarbonate can neutralize approximately 0.0498 g of stomach acid (HCl).

Now, let's move on to the second question about the production of ethanol from sucrose.

Sucrose (C12H22O11) is a disaccharide composed of glucose and fructose. The balanced chemical equation for the fermentation of sucrose to ethanol and carbon dioxide is as follows:

C12H22O11 -> 4C2H5OH + 4CO2

From the balanced equation, we can observe that one mole of sucrose produces four moles of ethanol. To find the mass of ethanol obtained from 6.00 kg of sucrose, we need to follow these steps:

Step 1: Convert the mass of sucrose from kg to grams:
Mass of sucrose = 6.00 kg x 1000 g/kg
= 6000 g

Step 2: Convert the mass of sucrose to moles using its molar mass:
Molar mass of sucrose = (12.01 x 12) + (1.01 x 22) + (16.00 x 11)
= 144.13 g/mol

Moles of sucrose = Mass of sucrose / Molar mass of sucrose
= 6000 g / 144.13 g/mol
= 41.62 mol

Step 3: Determine the moles of ethanol produced from the moles of sucrose:
Moles of ethanol = 4 x Moles of sucrose
= 4 x 41.62 mol
= 166.48 mol

Step 4: Convert moles of ethanol to mass using its molar mass:
Molar mass of ethanol = (12.01 x 2) + (1.01 x 5) + (16.00 x 1)
= 46.07 g/mol

Mass of ethanol = Moles of ethanol x Molar mass of ethanol
= 166.48 mol x 46.07 g/mol
= 7660.48 g

Therefore, approximately 7660.48 g (or 7.66 kg) of ethanol can be obtained from 6.00 kg of sucrose.