The balanced chemical equation would be as follows:
<span>
Mg + O2 → MgO2
</span>We are given the amounts of the reactants. We need to determine first which one is the limiting reactant. We do as follows:
62.0 g Mg (1 mol / <span>24.31 g ) = 2.55 mol Mg
</span>55.5 g O2 ( 1 mol / 32 g ) = 1.74 mol O2 -----> <span>consumed completely and therefore the limiting reactant
2.55 mol - 1.74 mol O2( 1 mol Mg / 1 mol O2 ) = 0.81 mol Mg excess</span>
Answer:
Explanation:
Start by writing water as HOH
Ca(OH)2 +CO2 = CaCO3+ HOH
Next pay attention to the CO2 going to CO3
We need an oxygen.
Fortunately that is provided by the (OH)2
Now we have
Ca(OH)2 + CO2 ==> CaCO3 + HOH
and believe it or not, that is balanced as it is
The left side has 1 Ca. So does the right side
The Left side has 1 C. So does the right side.
The left side has 2 H. So does the right side
The left side has 2 oxygens (in Ca(OH2)) + 2 oxygens in CO2
So the equation is balanced.
Answer:
9.07 L
Explanation:
<em>Calculate the volume occupied at s.t.p by 6.89 g of NH₃ gas [H = 1.0, N = 14.0].</em>
Step 1: Given and required data
- Molar mass of NH₃ (M): 17.0 g/mol
Step 2: Calculate the moles (n) of NH₃
We will use the following epxression.
n = m / M
n = 6.89 g / (17.0 g/mol) = 0.405 mol
Step 3: Calculate the volume occupied by 0.405 moles of NH₃ at STP
At STP, 1 mole of NH₃ occupies 22.4 L (assuming ideal behavior).
0.405 mol × 22.4 L/1 mol = 9.07 L
