Answer:
4.08 grams
Explanation:
Essentially, we're looking for the mass of HCl that "matches" 3.26 grams of magnesium hydroxide.
First, convert 3.26 grams of 
into moles by dividing by the molar mass. The molar mass of is 24.3 + 16 * 2 + 1 * 2 = 58.3 g/mol. So, 3.26 grams is equal to:
3.26 g ÷ 58.3 g/mol = 0.0559 mol
Notice that from the chemical equation, magnesium hydroxide and hydrochloric acid (HCl) have a ratio of 1 to 2. In other words, for every 0.0559 moles of , there are 0.0559 * 2 = 0.112 moles of HCl.
Finally, convert moles of HCl to grams by multiplying 0.112 by the molar mass, which is 1 + 35.45 = 36.45 g/mol:
0.112 mol HCl * 36.45 g/mol = 4.08 g HCl
The answer is thus 4.08 grams.
<em>~ an aesthetics lover</em>
M(NaCl)=50.0 g
m(H₂O)=150.0 g
m(solution)=m(NaCl)+m(H₂O)
w(NaCl)=100m(NaCl)/m(solution)=100m(NaCl)/{m(NaCl)+m(H₂O)}
w(NaCl)=100*50.0/(50.0+150.0)=25%
<span>From the balanced equation:
4mol Fe will produce 2mol Fe2O3
Molar mass Fe = 55.847g/mol
16.7gFe = 16.7/55.847 = 0.3mol Fe
This will produce 0.15mol Fe2O3
Molar mass Fe2O3 = 159.6887 g/mol
0.15mol = 159.6887*0.15 = 23.95g Fe2O3 produced
Hope this helps</span>
160 g of SO3 are needed to make 400 g of 49% H2SO4.
<h3>How many grams of SO3 are required to prepare 400 g of 49% H2SO4?</h3>
The equation of the reaction for the formation of H2SO4 from SO3 is given below as follows:
1 mole of SO3 produces 1 mole of H2SO4
Molar mass of SO3 = 80 g/mol
Molar mass of H2SO4 = 98 g/mol
80 g of SO3 are required to produce 98 og 100%H2SO4
mass of SO3 required to produce 400 g of 100 %H2SO4 = 80/98 × 400 = 326.5 g of SO3
Mass of SO3 required to produce 49% of 400 g H2SO4 = 326.5 × 49% = 160 g
Therefore, 160 g of SO3 are needed to make 400 g of 49% H2SO4.
Learn more about mass and moles at: brainly.com/question/15374113
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