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3 years ago

What is the mass if 1.72 moles of sodium nitrate

Chemistry

2 answers:

alexgriva [62]3 years ago

8 0

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

Alex3 years ago

8 0

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

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Explanation:

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3 years ago

if you mixed 72.9g hydrochloric acid(aq) with 150g silver acetate(aq), what would be the limiting reagent?

horsena [70]

Answer:

Silver Acetate would be the Limiting Reagent.

Explanation:

The balance chemical equation for the given double displacement reaction is as;

HCl + AgC₂H₃O₂ → AgCl + HC₂H₃O₂

Step 1: Calculate Moles of Starting Materials:

Moles of HCl:

Moles = Mass / M.Mass

Moles = 72.9 g / 36.46

Moles = 1.99 moles

Moles of AgC₂H₃O₂:

Moles = 150 g / 166.91 g/mol

Moles = 0.898 moles

Step 2: Find out Limiting reagent as:

According to balance chemical equation.

1 mole of HCl reacts with = 1 mole of AgC₂H₃O₂

So,

1.99 moles of HCl will react with = X moles of AgC₂H₃O₂

Solving for X,

X = 1.99 mol × 1 mol / 1 mol

X = 1.99 mol of AgC₂H₃O₂

Hence, to completely consume 1.99 moles of Hydrochloric acid we will require 1.99 moles of Silver Acetate, But, we are provided with only 0.898 moles of Silver Acetate. This means Silver Acetate will consume first in the reaction therefore, it is the LIMITING REAGENT.

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