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CaHeK987 [17]
4 years ago
5

A student uses 0.10 M HCl to simulate the acid concentration in the stomach. What volume( in ml) of “stomach acid” reacts with a

tablet that contains 0.10 g of magnesium hydroxide?
Chemistry
1 answer:
Tpy6a [65]4 years ago
3 0

Given, concentration of HCl = 0.10 M

Mass of Mg(OH)₂ = 0.10 g

Molar mass of Mg(OH)₂ = 58.3197 g/mol

Moles  = mass /molar mass

Mole of Mg(OH)₂ = 0.10 g/ 58.3197 g/mol

Mole of Mg(OH)₂ = 1.71

Mg(OH)₂(s) + 2HCl(aq) → MgCl₂(aq) + 2H₂O(l)

According to the balanced equation 1 moles of Mg(OH)₂ reacts with 2 moles of HCl.

Therefore, 1.71  moles of Mg(OH)₂ reacts with (2 x 1.71) or 3.42 moles of HCl

Molarity = moles / volume in L

Volume of HCl = 3.42 moles of HCl / 0.10 M HCl

Volume of HCl = 34.2 L or 34200 ml

34200 ml of “stomach acid” reacts with a tablet that contains 0.10 g of magnesium hydroxide.

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5 0
3 years ago
A mixture 21.7 g NaCl 3.74 g kcl and 9.76 g licl how many moles of nacl are in this mixture
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Mass of NaCl = 21.7 g
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Let's assume that mixture has only given compounds and free of impurities.
Then, we can present this as a mole percentage.

mole % = (moles of desired substance / Total moles of the mixture) x 100%

Hence,
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mole % of NaCl = (moles of NaCl / Total moles of the mixture) x 100%
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