M₁=6.584 g
m₂=4,194 g
m(H₂O)=m₁-m₂
w(H₂O)=m(H₂O)/m₁
w(H₂O)=(m₁-m₂)/m₁
w(H₂O)=(6.584-4.194)/6.584=0.3630 (36.30%)
the percentage by mass of water in the hydrate 36.30
That would be a depression..
If the reaction started with 32g of Sulfur and 56 grams of Iron, then mass of produced FeS is 88 grams.
<h3>What is the relation between mass & moles?</h3>
Relation between the mass and moles of any substance will be represented as:
n = W/M, where
- W = given mass
- M = molar mass
Moles of Fe = 56g / 56g/mol = 1 mol
Moles of S = 32g / 32g/mol = 1 mol
Given chemical reaction is:
Fe + S → FeS
From the stoichiometry of the reaction same moles of iron and sulfur are required to produce equal moles of iron sulfide. Means 1 mole of iron produces 1 moles of iron sulfide and mass of iron sulfide will be calculated as:
W = (1mol)(88g/mol) = 88g
Hence required mass of iron sulfide is 88 grams.
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Annual <span> $105,200 but yealy $135,670
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First we need to find the number of moles that 43.9g of gallium metal is. We can do this by finding the molar weight of gallium and cross-multiplying to cancel out units:
So we are dealing with 0.63 moles of gallium metal.
We can take from the balanced equation that 4 moles of gallium metal will react completely with 3 moles of oxygen gas. We can take this ratio and make a proportion to find the amount of oxygen gas, in moles, that will react completely with 0.63 moles of gallium metal:
Cross multiply and solve for x:
So now we know that 0.47 moles of oxygen gas will react with 43.9g of gallium metal.