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Vanyuwa [196]
3 years ago
11

What volume of 3.00 M HCl in liters is needed to react completely (with nothing left over) with 0.750 L of 0.100 M Na2CO3?

Chemistry
1 answer:
Alla [95]3 years ago
5 0

Answer:

0.0500L of 3.00M HCl are needed to react completely

Explanation:

Hydrochloric acid HCl reacts with sodium carbonate, Na2CO3, as follows:

2HCl + Na₂CO₃ → 2NaCl + H₂O + CO₂

<em>Where 2 moles of HCl reacts per mole of Na₂CO₃</em>

<em />

To find the volume of 3.00M HCl needed to react completely we need first to calculate the moles of Na₂CO₃ added and, with this chemical equation we can solve for moles of HCl as follows:

<em>Moles Na₂CO₃:</em>

0.750L * (0.100moles Na₂CO₃ / L) = 0.0750moles

<em>Moles HCl:</em>

0.0750moles Na₂CO₃ * (2 moles HCl / 1 mole of Na₂CO₃) = 0.150 moles of HCl

<em>Volume of 3.00M HCl:</em>

0.150 moles of HCl * (1L / 3.00moles) =

<h3>0.0500L of 3.00M HCl are needed to react completely</h3>
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3 0
3 years ago
What mass of sodium chloride (NaCl) forms when 7.5 g of sodium carbonate (Na2CO3) reacts with a dilute solution of hydrochloric
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 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

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3 years ago
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What is that the theoretical yield of aluminum oxide I if 3.20 mol of aluminum metal is exposed to 2.70 mole of oxygen
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Answer:

163.2g

Explanation:

First let us generate a balanced equation for the reaction. This is shown below:

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From the question given, were were told that 3.2moles of aluminium was exposed to 2.7moles of oxygen. Judging by this, oxygen is excess.

From the equation,

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Therefore, 3.2moles of Al will produce = (3.2x2)/4 = 1.6mol of Al2O3.

Now, let us covert 1.6mol of Al2O3 to obtain the theoretical yield. This is illustrated below:

Mole of Al2O3 = 1.6mole

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Number of mole = Mass /Molar Mass

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Mass of Al2O3 = 1.6 x 102 = 163.2g

Therefore the theoretical of Al2O3 is 163.2g

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