Answer:
D. (16.0 g + 16.0 g) × 100% / (32.1 g + 16.0 g + 16.0 g) = 49.9%
Explanation:
Step 1: Detemine the mass of O in SO₂
There are 2 atoms of O in 1 molecule of SO₂. Then,
m(O) = 2 × 16.0 g = 16.0 g + 16.0 g = 32.0 g
Step 2: Determine the mass of SO₂
m(SO₂) = 1 × mS + 2 × mO = 1 × 32.1 g + 2 × 16.0 g = 32.1 g + 16.0 g + 16.0 g = 64.1 g
Step 3: Detemine the mass percent of oxygen in SO₂
We will use the following expression.
m(O)/m(SO₂) × 100%
(16.0 g + 16.0 g) × 100% / (32.1 g + 16.0 g + 16.0 g) = 49.9%
Through the process of conduction, heat van be trnafered between two solids.
Answer:
2KBr + MgF₂ –> 2KF + MgBr₂
The coefficients are: 2, 1, 2, 1
Explanation:
KBr + MgF₂ –> KF + MgBr₂
The above equation can be balance as illustrated below:
KBr + MgF₂ –> KF + MgBr₂
There are 2 atoms of F on the left side and 1 atom on the right. It can be balance by writing 2 before KF as shown below:
KBr + MgF₂ –> 2KF + MgBr₂
There 2 atoms of K on the right side and 1 atom on the left side. It can be balance by writing 2 before KBr as shown below:
2KBr + MgF₂ –> 2KF + MgBr₂
Now, the equation is balanced.
The coefficients are: 2, 1, 2, 1
Answer:
0.7561 g.
Explanation:
- The hydrogen than can be prepared from Al according to the balanced equation:
<em>2Al + 6HCl → 2AlCl₃ + 3H₂,</em>
It is clear that 2.0 moles of Al react with 6.0 mole of HCl to produce 2.0 moles of AlCl₃ and 3.0 mole of H₂.
- Firstly, we need to calculate the no. of moles of (6.8 g) of Al:
no. of moles of Al = mass/atomic mass = (6.8 g)/(26.98 g/mol) = 0.252 mol.
<em>Using cross multiplication:</em>
2.0 mol of Al produce → 3.0 mol of H₂, from stichiometry.
0.252 mol of Al need to react → ??? mol of H₂.
∴ the no. of moles of H₂ that can be prepared from 6.80 g of aluminum = (3.0 mol)(0.252 mol)/(2.0 mol) = 0.3781 mol.
- Now, we can get the mass of H₂ that can be prepared from 6.80 g of aluminum:
mass of H₂ = (no. of moles)(molar mass) = (0.3781 mol)(2.0 g/mol) = 0.7561 g.
