1. If 23.00 grams of carbon dioxide and 18.75 grams of water are formed from the combustion reaction. What is the percent yield for CO2 and H20. How much in grams of the excess reactant remains?

2. Balance the equation and determine the mass in grams of hydrogen gas produced when 3 moles of tungsten react with excess oxalic acid to produce tungsten oxalate and hydrogen gas

1. What combustion reaction?

2. W + 2H2C2O4 ==> W(C2O4)2 + 2H2
Note: I would not think the above is the correct formula for tungsten oxalate but this was the only reference I could find.
3 mol W x (2 mol H2/1 mol W) = 6 mols H2 gas.
mass = mols H2 x molar mass H2.

To find the percent yield for CO2 and H2O in question 1, you need to compare the actual yield to the theoretical yield. The theoretical yield is the amount of product that would be formed if the reaction went to completion, and the actual yield is the amount of product actually obtained in the experiment.

To calculate the theoretical yield of CO2 and H2O, you first need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed in the reaction, which then determines the maximum amount of product that can be formed. To find the limiting reactant, you can use the concept of stoichiometry.

1. Calculate the amount of CO2 and H2O that would be formed from the given mass of carbon dioxide (23.00 grams) and water (18.75 grams). You can use the molar mass of CO2 (44.01 g/mol) and H2O (18.02 g/mol) to convert the given mass into moles:

- Moles of CO2 = 23.00 g / 44.01 g/mol
- Moles of H2O = 18.75 g / 18.02 g/mol

2. Next, you need to determine the mole ratio between the reactants and the products from the balanced chemical equation of the combustion reaction. The balanced equation will give you the stoichiometric ratios between the reactants and the products.

For example, if the balanced equation is:

C + O2 -> CO2 + H2O

The mole ratio between CO2 and C is 1:1, meaning one mole of C produces one mole of CO2. Similarly, the mole ratio between H2O and C is 1:1 as well.

3. Calculate the moles of CO2 and H2O that would be formed assuming complete reaction using the mole ratios from the balanced equation:

- Moles of CO2 = moles of C
- Moles of H2O = moles of C

4. Compare the moles of products calculated in step 3 to the actual moles obtained in step 1. The smaller value between the moles of CO2 and H2O will be the limiting reactant.

5. Once you have identified the limiting reactant, you can calculate the theoretical yield of CO2 and H2O based on the mole ratio from the balanced equation. Multiply the moles of the limiting reactant by the mole ratio between the product and limiting reactant to obtain the theoretical yield.

To calculate the percent yield of CO2 and H2O, use the following equation:

Percent Yield = (Actual Yield / Theoretical Yield) * 100

The actual yield is the given mass of CO2 and H2O (23.00 grams and 18.75 grams, respectively), and the theoretical yield can be calculated as explained above.

To find the grams of the excess reactant that remains, subtract the amount of the limiting reactant consumed from the initial amount of excess reactant.

For question 2, to balance the chemical equation and determine the mass of hydrogen gas produced, you need to follow these steps:

1. Write the unbalanced chemical equation based on the given reactants and products.

For example:

tungsten + oxalic acid -> tungsten oxalate + hydrogen gas

2. Balance the equation by adjusting the coefficients in front of each compound. Ensure that the number of atoms of each element is the same on both sides of the equation.

3. Once the equation is balanced, you can determine the mole ratio between the reactants and the product (hydrogen gas). This mole ratio can be obtained from the coefficients of the balanced equation.

4. Use the given moles of tungsten (in this case, 3 moles) to calculate the moles of hydrogen gas produced. Multiply the given moles of tungsten by the mole ratio between the product and tungsten.

5. Finally, convert the moles of hydrogen gas to grams by multiplying by the molar mass of hydrogen gas (2.016 g/mol).

Remember to always balance the equation before doing any calculations to ensure accurate results.