How many grams of P are present in 3.89 grams of tetraphosphorus decaoxide? How many grams of tetraphosphorus decaoxide contain 4.96 grams of O?

How many grams of P are present in 3.89 grams of tetraphosphorus decaoxide? How many grams of tetraphosphorus decaoxide contain 4.96 grams of O?

It’s right bc 3.89 is that

Answer:

87.71 grams

Explanation:

The molar mass of P4O10 is calculated as follows:

(4 atoms of P) × (atomic mass of P) + (10 atoms of O) × (atomic mass of O)

= (4 × 30.97 g/mol) + (10 × 16.00 g/mol)

= 123.88 g/mol + 160.00 g/mol

= 283.88 g/mol

Now, we can calculate the grams of P in 3.89 grams of P4O10 using the proportion:

(grams of P) = (grams of P4O10) × (molar mass of P) / (molar mass of P4O10)

= 3.89 g × (4 × 30.97 g/mol) / 283.88 g/mol

= 0.4268 g

Therefore, there are approximately 0.4268 grams of P in 3.89 grams of tetraphosphorus decaoxide.

To determine how many grams of tetraphosphorus decaoxide (P4O10) contain 4.96 grams of O (oxygen), we need to calculate the proportion of O in P4O10.

Using the molar mass of O (16.00 g/mol) and the molar mass of P4O10 (283.88 g/mol), we can set up the proportion:

(grams of P4O10) = (grams of O) × (molar mass of P4O10) / (molar mass of O)

= 4.96 g × 283.88 g/mol / 16.00 g/mol

= 87.71 g

Therefore, approximately 87.71 grams of tetraphosphorus decaoxide contain 4.96 grams of O.