Consider the following reaction:10N2(g)   +    25O2(g)    →    10N2O5(s)If 100g of O2 is reacted with 250g of N2, calculate the mass of solid N2O5 formed.Group of answer choices845g964 g135g193 g960 g

To solve this problem, we need to use stoichiometry. Here are the steps:

1. First, we need to calculate the molar mass of N2, O2, and N2O5. The molar mass of N2 is 28 g/mol, O2 is 32 g/mol, and N2O5 is 108 g/mol.

2. Next, we calculate the number of moles for N2 and O2. For N2, it's 250g / 28 g/mol = 8.93 mol. For O2, it's 100g / 32 g/mol = 3.125 mol.

3. According to the balanced equation, the mole ratio of N2:O2:N2O5 is 10:25:10. This means for every 10 moles of N2, we need 25 moles of O2 to produce 10 moles of N2O5.

4. We can see that we have less O2 (3.125 mol) than required (for 8.93 mol of N2, we would need 22.325 mol of O2). Therefore, O2 is the limiting reactant.

5. Now, we can calculate the amount of N2O5 produced. From the balanced equation, we know that 25 moles of O2 produce 10 moles of N2O5. So, 3.125 moles of O2 will produce (10/25)*3.125 = 1.25 moles of N2O5.

6. Finally, we convert the moles of N2O5 to grams. 1.25 moles * 108 g/mol = 135 g.

So, the mass of solid N2O5 formed is 135g.