Tin metal reacts with hydrogen fluoride to produce tin(II) fluoride and hydrogen gas according to the following balanced equation.
Sn(s)+2HF(g)→SnF2(s)+H2(g)
Sn(s)+2HF(g)→
SnF
2
(s)+
H
2
(g)
How many moles of hydrogen fluoride are required to react completely with 75.0 g of tin?
Step 1: List the known quantities and plan the problem.
Known
given: 75.0 g Sn
molar mass of Sn = 118.69 g/mol
1 mol Sn = 2 mol HF (mole ratio)
Unknown
mol HF
Use the molar mass of Sn to convert the grams of Sn to moles. Then use the mole ratio to convert from mol Sn to mol HF. This will be done in a single two-step calculation.
g Sn → mol Sn → mol HF
Step 2: Solve.
75.0 g Sn×1 mol Sn118.69 g Sn×2 mol HF1 mol Sn=1.26 mol HF
75.0 g Sn×
1
mol Sn
118.69
g Sn
×
2
mol HF
1
mol Sn
=1.26 mol HF
Step 3: Think about your result.
The mass of tin is less than one mole, but the 1:2 ratio means that more than one mole of HF is required for the reaction. The answer has three significant figures because the given mass has three significant figures.
Answer:
Magnesium
0.003mole
Explanation:
The problem here entails we find the metal in the carbonate.
For group 2 member, let the metal = X;
The carbonate is XCO₃;
If we sum the atomic mass of the elements in the metal carbonate, we should arrive at 84g/mol
Atomic mass of C = 12g/mol
O = 16g/mol
Atomic mass of X + 12 + 3(16) = 84
Atomic mass of X = 84 - 60 = 24g/mol
The element with atomic mass of 24g is Magnesium
B.
Number of moles in 0.3g of CaCO₃:
Molar mass of CaCO₃ = 40 + 12 + 3(16) = 100g/mol
Number of moles =
Number of moles =
= 0.003mole
Taking into account the definition of avogadro's number, 3.82×10⁻³ moles of H are 2.3×10²¹ particles of H.
<h3>
Avogadro's Number</h3>
Avogadro's Number or Avogadro's Constant is called the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of said substance. Its value is 6.023×10²³ particles per mole. Avogadro's number applies to any substance.
<h3>This case</h3>
Then you can apply the following rule of three: if 6.023×10²³ particles are contained in 1 mole of H, then 2.3×10²¹ particles are contained in how many moles of H?
amount of moles of H= (2.3×10²¹ particles × 1 mole)÷ 6.023×10²³ particles
<u><em>amount of moles of H= 3.82×10⁻³ moles</em></u>
Finally, 3.82×10⁻³ moles of H are 2.3×10²¹ particles of H.
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