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3 years ago

Calculate the amount in grams of Na2CO3 needed in a reaction with HCl to produce 120g NaCl?

1 answer:

8 0

The balanced chemical reaction is written as :

Na2CO3<span> + 2HCl === 2NaCl + H2O + CO2
</span>
We are given the amount of NaCl to be produced from the reaction. This will be the starting point for the calculations. We do as follows:

120 g NaCl ( 1 mol / 58.44 g) ( 1 mol Na2CO3 / 2 mol NaCl)( 105.99 g / 1 mol ) = 1108.82 g Na2CO3 needed

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What sample size (grams) of Na3PO4 (FW 164.00) known to be 50.00% pure should be used to consume exactly 40.00 mL of 0.1000 M HC

Mkey [24]

Answer:

0.109 g.

Explanation:

Equation of the reaction:

Na3PO4 + 3HCl --> 3NaCl + H3PO4

Number of moles of HCl = molar concentration × volume

= 0.1 × 0.04

= 0.004 mol.

By stoichiometry, 1 mole of Na3PO4 neutralises 3 moles of HCl. Therefore, number of moles of Na3PO4 = 0.004/3

= 0.0013 mol

Mass of Na3PO4 = molar mass × number of moles

= 0.0013 × 164

= 0.219 g

Since 50% of Na3PO4 was present in the sample. Let 100 g be the total mass of the substance

= 0.219 × 50 g/100 g

= 0.109 g.

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2 years ago

a 10.99g sample of NaBr contains 22.34% Na by mass. Considering the law of constant composition (define proportions), how many g

leonid [27]

Given :

A 10.99 g sample of NaBr contains 22.34% Na by mass.

To Find :

How many grams of sodium does a 9.77g sample of sodium bromine contain.

Solution :

By law of constant composition , in any given chemical compound, the elements always combine in the same proportion with each other.

Therefore , percentage of Na by mass in NaBr will be same for every amount .

Percentage of Na in 9.77 g NaBr is 22.34 % too .

Gram of Na = $9.77\times \dfrac{22.34}{100}=2.18\ g$ .

Hence , this is the required solution .

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Answer:

Explanation:

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