Question

For the reaction

2AgNO3+Na2CrO4⟶Ag2CrO4+2NaNO3

how many grams of silver chromate, Ag2CrO4, are produced from 69.1 g of silver nitrate, AgNO3?

Answer 1

66.4g

Step-by-step explanation:

Given parameters:

Mass of AgNO₃ = 69.1g

Unknown:

Mass of Ag₂CrO₄

Solution:

2AgNO₃+ Na₂CrO₄ ⟶ Ag₂CrO₄ + 2NaNO₃

The given equation is balanced.

To solve this problem, solve from the known to the unknown;

1. Find the number of moles of AgNO₃;

Number of moles =
`(mass)/(molar mass)`

atomic mass of Ag = 107.9g/mol

N = 14g/mol

O = 16g/mol

Molar mass of AgNO₃ = 107.9 + 14 + 3(16) = 169.9g/mol

Number of moles =
`(69.1)/(169.9)` = 0.41moles

2. From the balanced equation;

2 moles of AgNO₃ produced 1 moles of Ag₂CrO₄ ;

0.41 moles of AgNO₃ will produce
`(0.41)/(2)` = 0.2moles of Ag₂CrO₄

3:

Mass of Ag₂CrO₄ = number of moles x molar mass

Atomic mass of Ag = 107.9g/mol

Cr = 52g/mol

0 = 16g/mol

Molar mass = 2(107.9) + 52 + 4(16) = 331.8g/mol

so, mass of Ag₂CrO₄ = 0.2 x 331.8 = 66.4g