what mass of water is expected when 5.80 grams of sodium carbonate reacts with excess sulfuric acid? what volume of gas is produced at STP?

To find the mass of water produced when 5.80 grams of sodium carbonate reacts with excess sulfuric acid, we first need to determine the balanced chemical equation for the reaction.

The chemical equation for the reaction between sodium carbonate (Na2CO3) and sulfuric acid (H2SO4) can be written as:

Na2CO3 + H2SO4 → Na2SO4 + CO2 + H2O

From the equation, we can see that one mole of sodium carbonate reacts with one mole of sulfuric acid to produce one mole of water.

To calculate the mass of water, we need to convert the given mass of sodium carbonate to moles using the molar mass of sodium carbonate (Na2CO3).

The molar mass of Na2CO3 = (2 * atomic mass of Na) + atomic mass of C + (3 * atomic mass of O)
= (2 * 22.99 g/mol) + 12.01 g/mol + (3 * 16.00 g/mol)
= 105.99 g/mol

Now, let's calculate the number of moles of sodium carbonate:

Number of moles of Na2CO3 = Mass of Na2CO3 / Molar mass of Na2CO3
= 5.80 g / 105.99 g/mol
= 0.0547 mol

Since the reaction is balanced in a 1:1 ratio with water, the number of moles of water produced will also be 0.0547 mol.

To find the mass of water, we can use the molar mass of water (H2O), which is approximately 18.015 g/mol.

Mass of water = Number of moles of water * Molar mass of water
= 0.0547 mol * 18.015 g/mol
= 0.995 g (rounded to three significant figures)

Therefore, the expected mass of water produced when 5.80 grams of sodium carbonate reacts with excess sulfuric acid is approximately 0.995 grams.

To calculate the volume of gas produced at STP (Standard Temperature and Pressure), we can use the Ideal Gas Law.

The ideal gas law is expressed as:

PV = nRT

Where:
P = Pressure (in atm)
V = Volume of gas (in liters)
n = Number of moles
R = Ideal gas constant (0.0821 L.atm/mol.K)
T = Temperature (in Kelvin)

At STP, the pressure is 1 atm and the temperature is 273.15 K.

Since we know the number of moles of CO2 produced from the balanced equation (1 mole), we can directly calculate the volume.

V = nRT / P
= (1 mol) * (0.0821 L.atm/mol.K) * (273.15 K) / (1 atm)
= 22.414 L

Therefore, the volume of gas produced at STP is 22.414 liters.