Question

What is the mass if 1.72 moles of sodium nitrate

Answer 1

Answer:

146 g

Explanation:

Step 1. Calculate the molar mass of NaNO₃

Na =                    22.99

  N =                     14.01

3O = 3 × 16.00 = 48.00

               Total = 85.00 g/mol

Step 2. Calculate the mass  of NaNO₃

Mass of NaNO₃ = 1.72 × 85.00/1

Mass of NaNO₃ = 146 g

Answer 2

Answer:

146.2 g

Explanation:

What is the mass if 1.72 moles of sodium nitrate

The formula of sodium nitrate is NaNO3. First, we need to find the molar mass which is the sum of all atomic masses in the compound, in this case the atomic masses are:

Na = 22.99 g/mol

N = 14.01 g/mol

O = 16.00 g/mol

As we have 1 atom of Na, 1 atom of N and 3 atoms of O the formula to calculate the molar mass of sodium nitrate is:

Molar mass NaNO3 = 1( Na atomic mass) + 1(N atomic mass) + 3(O atomic mass)

Molar mass NaNO3 = 1(22.99 g/mol) + 1(14.01 g/mol) + 3(16.00 g/mol) = 85.00 g/mol

That is to say that a mole of sodium nitrate has a mass of 85.00 g

Then, by a rule of three we have:

1 mole NaNO3        ----- 85.00 g

1.72 moles NaNO3 -----     x

$x = \frac{(1.72 moles)(85.00g)}{1 mole} = 146.2 g$

So, the mass of 1.72 moles of sodium nitrate is 146.2 g