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LiRa [457]
2 years ago
5

in the reaction Mg (s) + 2HCl (aq) H2 (g) + MgCl2 (aq), how many moles of hydrogen gas will e produced from 75.0 milliliters of

a 1.0 M HCl in an excess of Mg?
Chemistry
1 answer:
Anna35 [415]2 years ago
6 0
This question uses mole ratios. firstly, we look at how many moles of H2 are produced for every mole of HCl: from 1 Hcl, we get half a mole of H2. therefore, however many moles of HCl are present in the 0.075 L of 1.0M HCl, half that number of H2 moles will be produced. 

n=cv
  =0.0750 x 1.0
  =0.075 mol HCl
n(H2)= 0.5 x n(HCl)
         =0.0375 moles of H2
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Answer:

\boxed {\boxed {\sf About \ 2.127 \ moles \ of \ FeCl_3}}

Explanation:

To convert from moles to grams, the molar mass must be used.

1. Find Molar Mass

The compound is iron (III) chloride: FeCl₃

First, find the molar masses of the individual elements in the compound: iron (Fe) and chlorine (Cl).

  • Fe:  55.84 g/mol
  • Cl:  35.45 g/mol

There are 3 atoms of chlorine, denoted by the subscript after Cl. Multiply the molar mass of chlorine by 3 and add iron's molar mass.

  • FeCl₃: 3(35.45 g/mol)+(55.84 g/mol)=162.19 g/mol

This number tells us the grams of FeCl₃ in 1 mole.

2. Calculate Moles

Use the number as a ratio.

\frac{162.19 \ g \ FeCl_3}{1 \ mol \ FeCl_3}

Multiply by the given number of grams, 345.0.

345.0 \ g \ FeCl_3 *\frac{162.19 \ g \ FeCl_3}{1 \ mol \ FeCl_3}

Flip the fraction so the grams of FeCl₃ will cancel.

345.0 \ g \ FeCl_3 *\frac{1 \ mol \ FeCl_3}{162.19 \ g \ FeCl_3}

345.0 *\frac{1 \ mol \ FeCl_3}{162.19 }

\frac{345.0 \ mol \ FeCl_3}{162.19 }

Divide.

2.12713484 \ mol \ FeCl_3

3. Round

The original measurement of grams, 345.0, has 4 significant figures. We must round our answer to 4 sig figs.

For the answer we calculated, that is the thousandth place.

The 1 in the ten thousandth place tells us to leave the 7 in the thousandth place.

\approx 2.127 \ mol \ FeCl_3

There are about <u>2.127 mole</u>s of iron (III) chloride in 345.0 grams.

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