Question

How many grams of zns are required to react with 12.6 l of oxygen gas at stp?

Answer 1

To determine the grams of ZnS required to react with 12.6 L of oxygen gas at STP, use the balanced chemical equation and the ideal gas law to calculate the number of moles of oxygen gas. Then, use the balanced equation again to calculate the number of moles of ZnS. Finally, convert moles of ZnS to grams using its molar mass.

Step-by-step explanation:

To determine the grams of ZnS required to react with 12.6 L of oxygen gas at STP, we need to first balance the chemical equation for the reaction between ZnS and oxygen.

The balanced equation is:

ZnS + 1.5 O2 -> ZnSO4

The molar ratio between ZnS and O2 is 1:1.5.

Therefore, we can calculate the number of moles of O2 using the ideal gas law: PV = nRT.

At STP, the pressure (P) is 1 atm and the volume (V) is 12.6 L.

The gas constant (R) is 0.0821 L·atm/mol·K, and the temperature (T) is 273 K.

Substituting these values into the ideal gas law equation, we can solve for the number of moles of O2.

With the number of moles of O2 known, we can use the balanced equation to determine the number of moles of ZnS. Finally, we can convert moles of ZnS to grams using its molar mass.

Answer 2

The balanced chemical equation:
2 ZnS + 3 O₂ → 2 ZnO + 2 SO₂

1 mole of any gas at STP occupies 22.4 L
?? mole of O₂ gas at STP will occupy 12.6 L

we have 0.5625 mole of O₂

From the balanced equation we see that:
2 moles of ZnS react with 3 moles O₂
?? mole ZnS react with 0.5625 mole O₂

Number of moles of ZnS = 0.375 mol
mass of ZnS = 0.375 * 97.45 (Molar mass) = 36.5 g ZnS