Answer:
Option B. 2Mg(s) + O2 (g) —> 2MgO (s)
Explanation:
From the question given above,
We were told that:
2 solid Mg atoms bond with O2 gas to produce solid MgO.
This can be represented by an equation as follow:
2Mg(s) + O2 (g) —> MgO (s)
Next, we shall balance the above equation as follow:
2Mg(s) + O2 (g) —> MgO (s)
There are 2 atoms of Mg on the left side and 1 atom on the right side. It can be balance by putting 2 in front of MgO as shown below:
2Mg(s) + O2 (g) —> 2MgO (s)
Now, the equation is balanced.
23.9 11.00 would be the wrong answer because I guessed
Answer:
132.17 g
Explanation:
The reaction given , in the question is -
CH₄ (g ) + 4 S ( g ) ---> CS₂ ( g ) + 2H₂S ( g )
From the reaction , 4 mole of S is required for the production of 1 mole of CS₂ .
since ,
Moles of CS₂ = given mass of CS₂ / Molecular weight of CS₂
Since ,
the Molecular weight of CS₂ = 76
Given , mass of CS₂ = 72.57 g
Moles of CS₂ = 72.57 / 76 = 0.95 mol
Since ,
The yield is 92.0 % .
Moles of S required = 4 * 0.95 mol / 0.92 = 4.13 moles
Mass of S required = 4.13 * 32 = 132.17 g .
The volume of a gas that is required yo react with 4.03 g mg at STP is 1856 ml
calculation/
- calculate the moles of Mg used
moles=mass/molar mass
moles of Mg is therefore=4.03 g/ 24.3 g/mol=0.1658 moles
- by use of mole ratio of Mg:O2 from the equation which is 2:1
the moles 02=0.1679 x1/20.0829 moles
- at STP 1 mole of a gas= 22.4 l
0.0895 moles=? L
- =0.0895 moles x22.4 l/ 1 mole=1.8570 L
into Ml = 1.8570 x1000=1856 ml approximately to 1860
Answer:CO2(g) will be formed at a faster rate in experiment 2 because more H+ particles can react per unit time
Explanation:
