Question

h₂ + o₂ → h₂o
if 4.0 moles of hydrogen gas reacts with 42g of oxygen gas, hov grams of water will be produced?
your answer
this is a required question
what is the percent by mass of oxygen in carbon dioxide? *
your answer
nitrogen gas and hydrogen gas react to form ammonia (equatior balanced):
n₂ + h₂ → nh₃
if 4.50 moles of n₂ reacts, how many moles of nh₃ will form?
your answer
this is a required question

Explanation:

First Question:

Step1: Balance the chemical equation

$$2H_{2}+O_{2} ightarrow 2H_{2}O$$

Step2: Calculate the moles of \(O_{2}\)

The molar mass of \(O_{2}\) is \(M = 32g/mol\). Using the formula \(n=\frac{m}{M}\), for \(m = 42g\) of \(O_{2}\), \(n_{O_{2}}=\frac{42g}{32g/mol}=1.3125mol\)

Step3: Determine the limiting reactant

From the balanced equation, the mole ratio of \(H_{2}\) to \(O_{2}\) is \(2:1\). If we have \(n_{H_{2}} = 4.0mol\), the moles of \(O_{2}\) required for complete reaction of \(H_{2}\) is \(n_{O_{2}\text{(required)}}=\frac{4.0mol}{2}=2.0mol\). Since we have \(n_{O_{2}}=1.3125mol<2.0mol\), \(O_{2}\) is the limiting reactant.

Step4: Calculate the moles of \(H_{2}O\) produced

From the balanced equation, the mole ratio of \(O_{2}\) to \(H_{2}O\) is \(1:2\). So \(n_{H_{2}O}=2\times n_{O_{2}}\). Substituting \(n_{O_{2}} = 1.3125mol\), \(n_{H_{2}O}=2\times1.3125mol = 2.625mol\)

Step5: Calculate the mass of \(H_{2}O\)

The molar mass of \(H_{2}O\) is \(M = 18g/mol\). Using \(m=n\times M\), \(m_{H_{2}O}=2.625mol\times18g/mol = 47.25g\)

Step1: Find the molar mass of \(CO_{2}\)

The molar mass of \(C\) is \(12g/mol\) and of \(O\) is \(16g/mol\). For \(CO_{2}\), \(M = 12g/mol+2\times16g/mol=44g/mol\)

Step2: Calculate the mass of \(O\) in \(CO_{2}\)

In \(CO_{2}\), the mass of \(O\) is \(2\times16g/mol = 32g/mol\) (per mole of \(CO_{2}\))

Step3: Calculate the mass - percent of \(O\)

Using the formula \(\text{Mass}\%=\frac{\text{Mass of component}}{\text{Mass of compound}}\times100\%\), \(\text{Mass}\%=\frac{32g/mol}{44g/mol}\times100\%=\frac{800}{11}\%\approx72.7\%\)

Step1: Balance the chemical equation

$$N_{2}+3H_{2} ightarrow 2NH_{3}$$

Step2: Use the mole ratio

From the balanced equation, the mole ratio of \(N_{2}\) to \(NH_{3}\) is \(1:2\)

Step3: Calculate the moles of \(NH_{3}\)

If \(n_{N_{2}} = 4.50mol\), then \(n_{NH_{3}}=2\times n_{N_{2}}\)

Answer:

\(47.25g\)

Second Question: