Answer: 0.01 moles of SrCO₃ Explanation: In this excersise we need to propose the reaction: K₂CO₃ + Sr(NO₃)₂ → 2KNO₃ + SrCO₃ As we only have data about the potassium carbonate we assume the strontium nitrite as the excess reactant. 1 mol of K₂CO₃ react to 1 mol of Sr(NO₃)₂ in order to produce 2 moles of potassium nitrite and 1 mol of strontium carbonate. Ratio is 1:1. In conclussion, 0.01 mol of K₂CO₃ must produce 0.01 moles of SrCO₃ Log in to Reply
Answer:
0.01 moles of SrCO₃
Explanation:
In this excersise we need to propose the reaction:
K₂CO₃ + Sr(NO₃)₂ → 2KNO₃ + SrCO₃
As we only have data about the potassium carbonate we assume the strontium nitrite as the excess reactant.
1 mol of K₂CO₃ react to 1 mol of Sr(NO₃)₂ in order to produce 2 moles of potassium nitrite and 1 mol of strontium carbonate.
Ratio is 1:1. In conclussion,
0.01 mol of K₂CO₃ must produce 0.01 moles of SrCO₃