1.) 22.5 grams of Potassium nitrate react with Sodium Bromide to produce how many grams of
Potassium Bromide?
2.)Scandium (III) Sulfide reacts with solid Aluminum to produce 16.3 grams of Aluminum Sulfide,
how many grams of Scandium (III) Sulfide were used in the reaction?
in both cases, write the equation.
Figure how many moles of the limiting reactant you have
The equation will tell you how many moles or product to expect
convert that back to grams
For #2, just work backwards. The key is to change everything to moles, and make sure the equation is balanced right.
To find the grams of potassium bromide produced in the reaction between potassium nitrate and sodium bromide, you will need to determine the limiting reactant first.
1. Start by writing a balanced chemical equation for the reaction:
2 KNO3 + 2 NaBr → 2 KBr + Na2SO4 + O2
2. Determine the molar mass of the potassium bromide (KBr). The molar mass of K is 39.10 g/mol, and the molar mass of Br is 79.90 g/mol. Therefore, the molar mass of KBr is 39.10 + 79.90 = 119 g/mol.
3. Convert the given mass of potassium nitrate (22.5 g) to moles. To do this, divide the mass by the molar mass of potassium nitrate. The molar mass of KNO3 is 101.10 g/mol.
22.5 g KNO3 * (1 mol KNO3 / 101.10 g KNO3) = 0.222 mol KNO3
4. Convert the given mass of sodium bromide (not given) to moles, assuming it is the limiting reactant. Let's call this mass x grams. The molar mass of NaBr is 102.89 g/mol.
x g NaBr * (1 mol NaBr / 102.89 g NaBr) = x/102.89 mol NaBr
5. Since the balanced chemical equation shows a 1:1 molar ratio between KNO3 and KBr, the moles of KBr formed will also be 0.222 mol.
6. Now, use the molar mass of KBr (119 g/mol) to calculate the mass of potassium bromide formed.
0.222 mol KBr * (119 g KBr / 1 mol KBr) = 26.46 g KBr
Therefore, 22.5 grams of potassium nitrate react with sodium bromide to produce 26.46 grams of potassium bromide.
To find the grams of scandium (III) sulfide used in the reaction with aluminum, you can follow a similar process:
1. Write a balanced chemical equation for the reaction:
2 Al + 3 Sc2S3 → 2 Al2S3 + 3 Sc
2. Determine the molar mass of aluminum sulfide (Al2S3). The molar mass of Al is 26.98 g/mol, and the molar mass of S is 32.07 g/mol. Therefore, the molar mass of Al2S3 is 2 * 26.98 + 3 * 32.07 = 150.18 g/mol.
3. Convert the given mass of aluminum sulfide (16.3 g) to moles. To do this, divide the mass by the molar mass of aluminum sulfide.
16.3 g Al2S3 * (1 mol Al2S3 / 150.18 g Al2S3) = 0.108 mol Al2S3
4. Since the balanced chemical equation shows a 3:2 molar ratio between Sc2S3 and Al2S3, the moles of Sc2S3 used will be:
(0.108 mol Al2S3 * 3 mol Sc2S3) / 2 mol Al2S3 = 0.162 mol Sc2S3
5. Finally, convert the moles of scandium (III) sulfide to grams using its molar mass (not specified in the question).
0.162 mol Sc2S3 * (molar mass of Sc2S3 / 1 mol Sc2S3) = grams of Sc2S3 used
Without the molar mass of scandium (III) sulfide, it is not possible to determine the exact answer to the second question.