 Chemistry
4 June, 12:04

# For the reaction2AgNO3+Na2CrO4⟶Ag2CrO4+2NaNO32AgNO3+Na2CrO4⟶Ag2CrO4+2NaNO3how many grams of sodium chromate, Na2CrO4, are needed to react completely with12.112.1g of silver nitrate, AgNO3?

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Answers (1)
1. 4 June, 12:24
0
Mass of Na₂CrO₄ = 5.75 g

Explanation:

First of all we will write the balance chemical equation.

2AgNO₃ + Na₂CrO₄ → Ag₂CrO₄ + 2NaNO₃

Now we will calculate the moles of AgNO₃.

Number of moles = mass / molar mass

Molar mass of AgNO₃ = 107.87 + 14 + 3 * 16 = 169.87 g/mol

Number of moles = mass / molar mass

Number of moles = 12.1 g / 169.87 g/mol = 0.071 mol

Now we will compare the moles of AgNO₃ and Na₂CrO₄ from balance chemical equation.

AgNO₃ : Na₂CrO₄

2 : 1

0.071 : 1/2 * 0.071 = 0.0355

Now we will calculate the mass of Na₂CrO₄.

Molar mass of Na₂CrO₄ = 23*2 + 52 + 16*4 = 162 g/mol

Mass of Na₂CrO₄ = number of moles * molar mass

Mass of Na₂CrO₄ = 0.0355 mol * 162 g/mol

Mass of Na₂CrO₄ = 5.75 g
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