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tigry1 [53]
3 years ago
11

What mass of sodium chloride (NaCl) forms when 7.5 g of sodium carbonate (Na2CO3) reacts with a dilute solution of hydrochloric

acid (HCl)?
Chemistry
2 answers:
Gre4nikov [31]3 years ago
8 0

 The mass of NaCl  formed  is 8.307  grams


<u><em> calculation</em></u>

step 1: write the equation  for reaction

Na₂CO₃  + 2HCl → 2 NaCl  +CO₂ +H₂O

Step 2: find the  moles of Na₂CO₃

moles = mass/molar mass

 The  molar mass of Na₂CO₃  is = (23 x2) + 12 + ( 16 x3) = 106 g/mol

moles  = 7.5 g/106 g/mol =0.071 moles

Step 3: use the  mole ratio to determine the  mole of NaCl

Na₂CO₃:NaCl  is  1:2  therefore the moles of NaCl =0.07  x2 =0.142 moles


Step 4:  calculate mass  of NaCl

mass= moles x molar mass

the molar  mass of NaCl= 23 +35.5 =58.5 g/mol

mass  = 0.142  moles x 58.5 g/mol =8.307  grams

Elena L [17]3 years ago
3 0

What mass of sodium chloride (NaCl) forms when 7.5 g of sodium carbonate (Na2CO3) reacts with a dilute solution of hydrochloric acid (HCl)?

Next, cancel like terms and units. Which of the following shows the correct terms and units canceled?

First Option

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2 years ago
How many grams of aluminum are required to react with 35 mL of 2.0 M hydrochloric acid, HCl?__ HCl + __ Al --&gt; __ AlCl3 + __
kiruha [24]

Answer:

0.6258 g

Explanation:

To determine the number grams of aluminum in the above reaction;

  • determine the number of moles of HCl
  • determine the mole ratio,
  • use the mole ratio to calculate the number of moles of aluminum.
  • use RFM of Aluminum to determine the grams required.

<u>Moles </u><u>of </u><u>HCl</u>

35 mL of 2.0 M HCl

2 moles of HCl is contained in 1000 mL

x moles of HCl is contained in 35 mL

x \: mol \:  =  \:  \frac{2 \:  \times  \: 35}{1000}  \\  = 0.07 \: moles \:

We have 0.07 moles of HCl.

<u>Mole </u><u>ratio</u>

  • Balanced equation

6HCl(aq) + 2Al(s) --> 2AlCl3(aq) + 3H2(g)

Hence mole ratio = 6 : 2 (HCl : Al

  • but moles of HCl is 0.07, therefore the moles of Al;

=  \frac{2}{6}  \times 0.07 \\  \:  = 0.0233333 \: moles

Therefore we have 0.0233333 moles of aluminum.

<u>Grams of </u><u>Aluminum</u>

We use the formula;

grams \:  = moles \:  \times  \: rfm

The RFM (Relative formula mass) of aluminum is 26.982g/mol.

Substitute values into the formula;

= 0.0233333 \: moles  \:  \times  \: 26.982 \:  \frac{g}{mol}  \\  = 0.625799 \: grams

The number of grams of aluminum required to react with HCl is 0.6258 g.

3 0
2 years ago
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