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Nat2105 [25]
3 years ago
14

How many milliliters of a 1.25 molar hydrochloric acid (HCl) solution would be needed to react completely with 60.0 grams of cal

cium metal? (2 points)
Ca (s) + 2HCl (aq)yields CaCl2 (aq) + H2 (g)
Chemistry
1 answer:
stich3 [128]3 years ago
4 0

<u>Answer:</u>

2400 mL

<u>Explanation:</u>

Ca + 2HCl \implies CaCl_2 + H_2

According to this equation, the stoichiometric ratio between Ca and HCl for the complete reaction is 1:2.

We know that the number of moles of Ca can be calculated using the mole formula. (<em>number of moles = mass / molar mass</em>)

Moles of Calcium = \frac{60}{40} = 1.5 mol

So the moles of HCl = 1.5 \times 2 = 3.0 mol

<em>Volume of HCl solution = Moles of HCl/ concentration of HCl</em>

Volume of HCl solution = \frac{3}{1.25} = 2400 mL

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0.052636002587839 moles

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Aqueous hydrochloric acid HCl will react with solid sodium hydroxide NaOH to produce aqueous sodium chloride NaCl and liquid wat
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Answer:

0.26g of NaCl is the maximum mass that could be produced

Explanation:

Based on the reaction:

HCl + NaOH → NaCl + H₂O

<em>Where 1 mol of HCl reacts per mol of NaOH to produce 1 mol of NaCl</em>

<em />

To solve this question we need to find <em>limiting reactant. </em>The moles of limiting reactant = Moles of NaCl produced:

<em>Moles HCl -Molar mass: 36.46g/mol-:</em>

0.365g HCl * (1mol / 36.46g) = 0.010 moles HCl

<em>Moles NaOH -Molar mass: 40g/mol-:</em>

0.18g NaOH * (1mol / 40g) = 0.0045 moles NaOH

As the reaction is 1:1 and moles NaOH < moles HCl, limiting reactant is NaOH and maximum moles produced of NaCl are 0.0045 moles.

The mass of NaCl is:

<em>Mass NaCl -Molar mass: 58.44g/mol-:</em>

0.0045 moles * (58.44g/mol) =

<h3>0.26g of NaCl is the maximum mass that could be produced</h3>
8 0
3 years ago
22 Determine the molar mass of an unknown gas that has a volume of 72.5 mL at a temperature of 68C, and a pressure of 0.980 atm
jekas [21]

Answer:

81.5g/mol

Explanation:

Molar mass is the ratio between mass of a substance (In this case, 0.207g) and moles presents in this mass.

To solve this question we must find the moles of the gas in order to obtain the molar mass using:

PV = nRT

PV / RT = n

<em>Where P is pressure = 0.980atm</em>

<em>V is volume in Liters = 0.0725L</em>

<em>R is gas constant = 0.082atmL/molK</em>

<em>T is absolute temperature = 68°C + 273.15 = 341.15K</em>

<em />

0.980atm*0.0725L / 0.082atmL/molK*341.15K = n

2.54x10⁻³ moles = n

Thus, the molar mass of the gas is:

0.207g / 2.54x10⁻³ moles

<h3>81.5g/mol</h3>
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Explanation:

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