Question

The equation for the reaction between calcium carbonate and dilute nitric acid is shown.
CaCO3(s) + 2HNO3(aq) → Ca(NO3)2(aq) + CO2(g) + H2O(l)
25 g of calcium carbonate is reacted with an excess of dilute nitric acid.
Which mass of calcium nitrate and which volume of carbon dioxide is produced at room
temperature and pressure?

Answer 1

The mass of calcium nitrate produced is 41 grams and volume of carbon dioxide produced is 5.6 L

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$     .....(1)

Given mass of calcium carbonate = 25 g

Molar mass of calcium carbonate = 100 g/mol

Putting values in equation 1, we get:

$\text{Moles of calcium carbonate}=\frac{25g}{100g/mol}=0.25mol$

The given chemical equation follows:

$CaCO_3(s)+2HNO_3(aq.)\rightarrow Ca(NO_3)_2(aq.)+CO_2(g)+H_2O(l)$

• For calcium nitrate:

By Stoichiometry of the reaction:

1 mole of calcium carbonate produces 1 mole of calcium nitrate

So, 0.25 moles of calcium carbonate will produce = $\frac{1}{1}\times 0.25=0.25mol$ of calcium nitrate

Now, calculating the mass of calcium nitrate by using equation 1:

Moles of calcium nitrate = 0.25 moles

Molar mass of calcium nitrate = 164 g/mol

Putting values in equation 1, we get:

$0.25mol=\frac{\text{Mass of calcium nitrate}}{164g/mol}\\\\\text{Mass of calcium nitrate}=(0.25mol\times 164g/mol)=41g$

• For oxygen gas:

By Stoichiometry of the reaction:

1 mole of calcium carbonate produces 1 mole of carbon dioxide

So, 0.25 moles of calcium carbonate will produce = $\frac{1}{1}\times 0.25=0.25mol$ of carbon dioxide

At STP:

1 mole of a gas occupies 22.4 L of volume

So, 0.25 moles of carbon dioxide gas will occupy = $\frac{22.4}{1}\times 0.25=5.6L$ of volume

Learn more about number of moles and volume at STP:

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