How many formula units of sodium nitrate will react with 20.0 g of calcium carbonate to
product sodium carbonate and calcium nitrate according to the following reactions?

2 na(no3) + ca(co3) → na2(co3) + ca(no3)2

2.41 × 10^23 formula units

The equation for the reaction is;

2 Na(NO₃) + Ca(CO₃) → Na₂(CO₃) + Ca(NO₃)₂

We are given;

Mass of calcium carbonate as 20.0 g

We need to determine the number of formula units of sodium nitrate that will react with 20 g of calcium carbonate.

To calculate the number of moles we divide mass by the molar mass;

Number of moles =  Mass ÷ Molar mass

Molar mass of CaCO₃ = 100.087 g/mol

Number of moles = 20 g ÷ 100.087 g/mol

                             = 0.20 moles

From the equation, 2 moles of NaNO₃ reacts with 1 mole of CaCO₃

Thus, the mole ratio of NaNO₃ to CaCO₃ is 2 : 1

Therefore; Number of moles of NaNO₃ = Moles of CaCO₃ × 2

                                                                 = 0.20 moles × 2

                                                                 = 0.40 moles

We need to know that 1 mole of a compound contains formula units equivalent to the Avogadro's constant.

That is; 1 mole of a compound contains 6.022 × 10^23 formula units

Therefore; 0.4 moles of NaNO₃ will have;

= 0.4 moles × 6.022 × 10^23 formula units

= 2.4088 × 10^23 formula units

= 2.41 × 10^23 formula units

Therefore, the number of formula units of sodium nitrate required will be 2.41 × 10^23 formula units

there all right good

explanation:

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