Step 1.
Find limiting agent (g of H2 / amu) and (g of Cl2 / amu)
Cl2 is less mole so it is limiting
Step 2.
Find the ratio (1:2)
1H + 1Cl = 2 HCl
Step 3
mole of limiting x ratio (2) x amu of HCl
Source
Mastering Chemistry
How many moles of NH3 are produced when 1.2 mol of nitrogen reacts with hydrogen?
How many moles of NH3 are produced when 1.2 mol of nitrogen reacts with hydrogen?
1,4 moles carbon monoxide are produced.
0,75 moles of AlCl3 (anhydrous) is equivalent to 100,005 g.
The reaction is:2 C + O2 = 2 COSo 2,1 moles are obtained.
Balanced equation: 2Al + 3Cl2 -> 2AlCl3For every 3 moles of Cl2, 2 moles of AlCl3 is produced (Using the numbers in front of the compounds)Now set up a proportion: 3/2 = 0.30/?Cross Multiply: (2 X 0.30) / 3 = 0.20.2 moles of AlCl3 will be produced.
How many moles of NH3 are produced when 1.2 mol of nitrogen reacts with hydrogen?
How many moles of NH3 are produced when 1.2 mol of nitrogen reacts with hydrogen?
1,4 moles carbon monoxide are produced.
The empirical formula for aluminum chloride is AlCl3, and its gram formula mass is 133.34. The formula shows that each formula unit contains one aluminum atom, and the the gram atomic mass of aluminum is 26.9815. Therefore, 18(133.34/26.9815) or 89 grams, to the justified number of significant digits, of aluminum chloride will be produced.
If 1 mole of carbon reacts, 1 mole of CO is produced according to the balanced chemical equation for the reaction. Therefore, if 1.4 moles of carbon react, 1.4 moles of CO will be produced.
0,75 moles of AlCl3 (anhydrous) is equivalent to 100,005 g.
The reaction is:2 C + O2 = 2 COSo 2,1 moles are obtained.
The balanced chemical equation for the reaction between aluminum and chlorine is 2Al + 3Cl2 -> 2AlCl3. This means that for every 2 moles of aluminum that react, 2 moles of aluminum chloride are produced. Therefore, if 0.440 mol of aluminum is used, it will produce 0.440 mol of aluminum chloride.
In 1 mol of AlCl3, there are 3 chloride ions. First calculate the moles of AlCl3 in the solution: 65.5 mL is 0.0655 L. Multiply 0.0655 L by 0.210 mol/L to get the moles of AlCl3. Finally, multiply this by 3 to find the number of chloride ions in the solution.
The balanced chemical equation for the reaction between H2 and NH3 is: 3H2 + N2 → 2NH3 From the equation, we can see that 3 moles of H2 produce 2 moles of NH3. Therefore, when 1.2 moles of H2 react, we can calculate the moles of NH3 produced as: 1.2 mol H2 * (2 mol NH3 / 3 mol H2) = 0.8 mol NH3.
3.03